U.S. patent application number 12/280739 was filed with the patent office on 2009-11-19 for energy storage and generation.
This patent application is currently assigned to HIGHVIEW ENTERPRISES LIMITED. Invention is credited to Ferdinand Berger, Haisheng Chen, Yulong Ding, Toby Peters.
Application Number | 20090282840 12/280739 |
Document ID | / |
Family ID | 38093272 |
Filed Date | 2009-11-19 |
United States Patent
Application |
20090282840 |
Kind Code |
A1 |
Chen; Haisheng ; et
al. |
November 19, 2009 |
ENERGY STORAGE AND GENERATION
Abstract
The present invention concerns systems for storing energy and
using the stored energy to generate electrical energy or drive a
propeller (505). In particular, the present invention provides a
method of storing energy comprising: providing a gaseous input,
producing a cryogen from the gaseous input; storing the cryogen;
expanding the cryogen; using the expanded cryogen to drive a
turbine (320) and recovering cold energy from the expansion of the
cryogen. The present invention also provides a cryogenic energy
storage system comprising: a source of cryogen; a cryogen storage
facility (370); means for expanding the cryogen; a turbine (320)
capable of being driven by the expanding cryogen; and means (340,
350) for recovering cold energy released during expansion of the
cryogen.
Inventors: |
Chen; Haisheng; (Beijing,
CN) ; Ding; Yulong; (Harrogate, GB) ; Peters;
Toby; (East Sussex, GB) ; Berger; Ferdinand;
(East Sussex, GB) |
Correspondence
Address: |
Stephen B. Salai, Esq.;Harter Secrest & Emery LLP
1600 Bausch & Lomb Place
Rochester
NY
14604-2711
US
|
Assignee: |
HIGHVIEW ENTERPRISES
LIMITED
East Sussex
GB
|
Family ID: |
38093272 |
Appl. No.: |
12/280739 |
Filed: |
February 27, 2007 |
PCT Filed: |
February 27, 2007 |
PCT NO: |
PCT/GB07/00667 |
371 Date: |
March 11, 2009 |
Current U.S.
Class: |
62/50.3 ; 60/651;
60/652; 60/671; 62/50.2 |
Current CPC
Class: |
F25J 1/0251 20130101;
F25J 2215/40 20130101; F25J 2245/40 20130101; F05B 2210/12
20130101; F25J 3/04496 20130101; Y02E 10/72 20130101; F25J 2230/20
20130101; F25J 2240/90 20130101; F25J 1/0045 20130101; F25J 3/04115
20130101; F03D 9/17 20160501; F17C 9/04 20130101; F25J 1/004
20130101; Y02E 60/16 20130101; F03D 9/25 20160501; F25J 3/04836
20130101; Y02E 70/30 20130101; F01D 15/04 20130101; F25J 1/0228
20130101; F25J 3/04533 20130101; F25J 1/0281 20130101; F25J 2205/62
20130101; F25J 3/04018 20130101; F02C 1/002 20130101; F25J 2235/02
20130101; F01K 15/04 20130101; F25J 3/04593 20130101; F25J 2230/06
20130101; F25J 2230/30 20130101; Y02T 50/60 20130101; F02C 1/05
20130101; F25J 3/04612 20130101; F02C 6/14 20130101; F25J 2240/10
20130101; F25J 1/0242 20130101; F25J 3/04618 20130101; F01K 3/00
20130101; F25J 1/0012 20130101; F25J 3/04224 20130101; F25J 2260/20
20130101; F25J 1/0234 20130101; F01D 15/005 20130101; F05D 2210/12
20130101; F01K 13/00 20130101; F25J 3/04078 20130101; F25J 1/0221
20130101; F25J 3/04169 20130101; F25J 2210/40 20130101 |
Class at
Publication: |
62/50.3 ;
62/50.2; 60/651; 60/652; 60/671 |
International
Class: |
F17C 9/04 20060101
F17C009/04; F02C 1/02 20060101 F02C001/02; F02C 6/14 20060101
F02C006/14 |
Foreign Application Data
Date |
Code |
Application Number |
Feb 27, 2006 |
GB |
0603895.4 |
May 5, 2006 |
GB |
0608959.3 |
Nov 3, 2006 |
GB |
0621972.9 |
Claims
1. A method of storing energy comprising: providing a gaseous
input; producing a cryogen from the gaseous input; storing the
cryogen; expanding the cryogen; using the expanded cryogen to drive
a turbine; and recovering cold energy from the expansion of the
cryogen.
2. The method of claim 1 further comprising using the turbine to
drive a generator and generate electricity.
3. The method of claim 1 further comprising using the turbine to
drive a propeller.
4. The method of claim 1 wherein the cryogen is liquid air.
5. The method of claim 4 wherein the cryogen is slush air.
6. The method of claim 1 wherein the step of expanding the cryogen
comprises heating the cryogen.
7. The method of claim 6 wherein the step of heating the cryogen
comprises heating the cryogen using ambient heat.
8. The method of claim 6 wherein the step of heating the cryogen
comprises heating the cryogen using geothermal heat.
9. The method of claim 6 wherein the step of heating the cryogen
comprises heating the cryogen using waste heat from a power plant,
from a steam stream, from the flue gas of a power-plant or from
another waste heat resource.
10. The method of claim 1 wherein the step of producing the cryogen
comprises compressing the gaseous input.
11. The method of claim 10 wherein the step of expanding the
cryogen comprises heating the cryogen using waste heat generated
during the step of compressing the gaseous input.
12. The method of claim 1 wherein the step of expanding the cryogen
comprises:. heating the cryogen to approximately the environmental
temperature using ambient air; then heating the cryogen further
using waste heat.
13. The method of claim 1 wherein the pressure of the cryogen is
increased prior to expansion.
14. The method of claim 1 wherein the cryogen is stored at an
increased temperature prior to expansion.
15. The method of claim 1 further comprising using the recovered
cold energy.
16. The method of claim 1 further comprising using the recovered
cold energy to enhance the production of more cryogen.
17. The method of claim 1 further comprising using the recovered
cold energy for refrigeration.
18. The method of claim 1 further comprising using the recovered
cold energy for air conditioning.
19. The method of claim 1 further comprising using waste heat
generated during the step of producing the cryogen to provide hot
air for heating.
20. The method of claim 1 further comprising using waste heat
generated during the step of producing the cryogen to provide hot
water.
21. The method of claim 1 wherein a non-polluting source of energy
is used to power the method.
22. The method of claim 1 further comprising the step of separating
contaminants from the gaseous input.
23. The method of claim 1 wherein the turbine comprises a
multi-stage quasi-isothermal turbine.
24. A cryogenic energy -storage system comprising: a source of
cryogen; a cryogen storage facility; means for expanding the
cryogen; a turbine capable of being driven by the expanding
cryogen; and means for recovering cold energy released during
expansion of the cryogen.
25. The cryogenic energy storage system of claim 24 further
comprising a generator wherein the generator is capable of being
driven by the turbine.
26. The cryogenic energy storage system of claim 24 further
comprising a propeller wherein the propeller is capable of being
driven by the turbine.
27. The cryogenic energy storage system of claim 24 wherein the
cryogen is liquid air.
28. The cryogenic energy storage system of claim 24 wherein the
cryogen is slush air.
29. The cryogenic energy storage system of claim 24 wherein the
source of cryogen is an air liquefaction plant.
30. The cryogenic energy storage system of claim 24, wherein the
means for expanding the cryogen comprises means for heating the
cryogen.
31. The cryogenic energy storage system of claim 30 wherein the
means for heating the cryogen comprises at least one heat
exchanger.
32. The cryogenic energy storage system of claim 31 wherein the at
least one heat exchanger is arranged to heat the cryogen using heat
from ambient air.
33. The cryogenic energy storage system of claim 31 wherein the at
least one heat exchanger is arranged to heat the cryogen using
geothermal heat.
34. The cryogenic energy storage system of claim 31 wherein the at
least one heat exchanger is arranged to heat the cryogen using
waste heat .from a power plant, from a steam stream, from the flue
gas of a power plant or from another waste heat resource.
35. The cryogenic energy storage system of claim 24, wherein the
source of cryogen comprises a source of gaseous input, a compressor
for compressing the gaseous input and at least one heat exchanger
for cooling the gaseous input.
36. The cryogenic energy storage system of claim 35 wherein the at
least one heat exchanger is arranged to heat and expand the cryogen
using waste heat generated during compression of the gaseous
input.
37. The cryogenic energy storage system of claim 35 wherein the at
least one heat exchanger is arranged to provide hot air for heating
using waste heat generated during compression of the gaseous
input.
38. The cryogenic energy storage system of claim 35 wherein the at
least one heat exchanger is arranged to provide hot water using
waste heat generated during compression of the gaseous input.
39. The cryogenic energy storage system of claim 31 comprising: a
first heat exchanger arranged to heat the cryogen to approximately
the environmental temperature using ambient air; and a second heat
exchanger arranged to heat the cryogen further using waste
heat.
40. The cryogenic energy storage system of claim 24 further
comprising a pump arranged to increase the pressure of the cryogen
prior to expansion.
41. The cryogenic energy storage system of claim 24 further
comprising a throttling valve arranged to convert the gaseous input
into a cryogen.
42. The cryogenic energy storage system of claim 31 wherein the
means for recovering cold energy from the expansion of the cryogen
comprises at least one heat exchanger arranged to use the cold
energy.
43. The cryogenic energy storage system of claim 31 wherein the
means for recovering cold energy from the expansion of the cryogen
comprises at least one heat exchanger arranged toy use the cold
energy to enhance the production of more cryogen.
44. The cryogenic energy storage system of claim 31 wherein the
means for recovering cold energy from the expansion of the cryogen
comprises at least one heat exchanger arranged to use the cold
energy for refrigeration.
45. The cryogenic energy storage system of claim 31 wherein the
means for recovering cold energy from the expansion of the -cryogen
comprises at least one heat exchanger arranged to use the cold
energy for air conditioning.
46. The cryogenic energy storage system of claim 22 further
comprising means for separating contaminants from the gaseous
input.
47. The cryogenic energy storage system of claim 22 wherein the
turbine comprises a multi-stage quasi-isothermal turbine.
48. (canceled)
49. (canceled)
50. (canceled)
Description
FIELD OF THE INVENTION
[0001] The present invention concerns systems for storing energy
and using the stored energy to generate electrical energy or drive
a propeller.
BACKGROUND TO THE INVENTION
[0002] Electrical energy storage systems store base-load energy
during off-peak periods and use the stored energy to provide
electrical power during peak periods. Such systems are essential to
the power generation industries. In conventional power generation
systems, an energy storage system can provide substantial benefits
including load following, peaking power and standby reserve. By
providing spinning reserve and a dispatched load, electrical energy
storage systems can increase the net efficiency of thermal power
sources while reducing harmful emissions.
[0003] Electrical energy storage systems are critically important
to intermittent renewable energy supply systems such as solar
photovoltaic and wind turbine supply systems. This is due to the
intermittent nature of the sources of renewable energy; the source
is not always available over an extended period of time. Such a
disadvantage has become an obstacle to the green electricity
industry. Therefore, there is a need for a suitable energy storage
system. Moreover, there is a need for the electricity storage
system to be green.
[0004] Furthermore, electrical energy storage systems are regarded
as a key technology in energy distribution networks with
distributed generators, in order to compensate for any power
fluctuation and to provide uninterruptible power supply during
periods of voltage drop due to, for example, line faults.
[0005] Several electrical energy storage systems have been
developed in the past. These include pumped hydro storage systems,
Compressed Air Energy Storage systems (CAES), secondary batteries,
Superconducting Magnetic Energy Storage systems (SMES), flywheels
and capacitors.
[0006] Pumped hydro is the most widely used form of energy storage
system. It stores hydraulic potential energy by pumping water from
a lower reservoir to a higher reservoir. The amount of stored
energy is proportional to the height difference between the two
reservoirs and the volume of water stored. During periods of high
demand for electricity, water falls from the higher reservoir to
the lower reservoir through a turbine generator in a manner similar
to traditional hydroelectric facilities. Pumped hydro storage is a
mature technology with high efficiency, large volume, long storage
period and relatively low capital cost per unit energy. However, a
scarcity of available sites for two large reservoirs and one or
more dams is the major drawback of pumped hydro. A long lead time
for construction (typically .about.10 years) and environmental
issues (e.g. removing trees and vegetation from the land prior to
the reservoir being flooded) are two other major drawbacks of the
pumped hydro system.
[0007] Compressed Air Energy Storage (CAES) is based on
conventional gas turbine technology. It uses the elastic potential
energy of compressed air. Energy is stored by compressing air in an
air tight space such as underground storage cavern. To extract the
stored energy, compressed air is drawn from the storage vessel,
heated and then expanded through a high pressure turbine, which
captures some of the energy in the compressed air. The air is then
mixed with fuel and combusted, with the exhaust expanded through a
low pressure turbine. Both the high and low pressure turbines are
connected to a generator to produce electricity. CAES has a
relatively high energy density, long storage period, low capital
costs and high efficiency. In comparison with pumped hydro and
other currently available energy storage systems, CAES is not an
independent system. It requires combustion in the gas turbine. It
cannot be used in other types of power plants such as coal-fired,
nuclear, wind turbine or solar photovoltaic plants. In addition,
the combustion of fossil fuels leads to emission of contaminates
such as nitrogen oxides and carbon oxides which render the CAES
less attractive. Also, similar to pumped hydro systems, CAES
suffers from a reliance on favourable geography such as caverns.
CAES can only be economically feasible for power plants that have
nearby rock mines, salt caverns, aquifers or depleted gas fields.
In addition, a major barrier for the CAES is the relatively low
pressures that can be achieved, typically 40-60 bar.
[0008] Secondary battery systems are in some ways ideally suited
for electrical energy storage systems. They not only provide fuel
flexibility and environmental benefits, but also offer a number of
important operating benefits to the electricity supply industry.
They can respond very rapidly to load changes, and they can accept
co-generated and/or third-party power, thus enhancing system
stability. The construction of a secondary battery system is
facilitated by short lead times, the lack of geographical
limitations on location, and the technology's modularity. However,
until recently, utility battery storage has been rare because of
the low energy densities, high maintenance costs, short lifetimes,
limited discharge capabilities and toxic remains associated with
such systems. There are several new battery technologies now
regarded as potentially competitive with pumped hydro and CAES
systems including lead acid batteries, sodium sulphur batteries,
zinc bromine batteries and redox flow batteries.
[0009] Superconducting Magnetic Energy Storage (SMES) is the only
known method for the bulk storage of energy directly as
electricity. SMES stores electrical energy as electric current
passing through an inductor. The inductor, made from a
superconducting material, is circular so that current can circulate
indefinitely with almost no losses. SMES exhibits very high energy
storage efficiency (typically .about.90%) and rapid respond (<1
second) relative to other energy storage systems. The major
problems confronting the implementation of SMES units are the high
cost and environmental issues associated with the strong magnetic
fields employed.
[0010] Flywheel systems are a form of energy storage system that
have been used for thousands of years. The disadvantages of these
systems are their short duration, relatively high frictional losses
(windage) and low energy densities. Traditional flywheel systems
with conventional metal rotors lack the necessary energy density to
be considered seriously for large-scale energy storage
applications. Recent advances in material science have started to
change this picture. In particular, the development of low-density,
high-strength, fibre-composite materials has allowed the design and
construction of flywheel energy storage systems with a comparable
energy density to other systems. Also, new bearing technologies are
being developed, such as levitation bearings using high temperature
superconductors which have, the potential of reducing the windage
losses that account for a large portion of the total energy
loss.
[0011] Capacitors are a form of energy storage system that have
been used for many years in the electronics industry. Double layer
capacitors have been developed for a daily peak load in the summer
of less than 1 hour with small capacities. Recent progress in the
field of redox super capacitors could lead to the development of
larger capacity systems. The major disadvantages of capacitors as
energy storage systems are, similar to flywheels, their short
duration and high energy dissipation due to self-discharge
loss.
[0012] Accordingly, there is a need for an electrical energy
storage system which has high energy density and potential output
power, high energy efficiency, a long duration, a long lifetime,
low capital costs, and offers a good commercial potential. The
system should preferably be capable of being used with current
power plants without requiring major modifications to the power
plants except to the inputs and outputs for electricity. The system
should also preferably be capable of working completely separately
from the power plant. Start-up and suspension of the system should
preferably be simple and reliable and the system should preferably
be capable of being used with most types of existing medium to
large scale power plants including coal-fired, gas turbine,
nuclear, wind turbine and solar photovoltaic plants, irrespective
of the geographical location of the plants. The system should also
preferably not be detrimental to the environment, particularly by
using the process in conjunction with non-polluting power plants (a
Zero Emission System), and may even have the potential to reverse
environmental impacts associated with the burning of fossil
fuels.
[0013] The inventors of the present invention have attempted to
provide an electrical energy storage system that addresses these
requirements.
[0014] In addition, there is also a need for an improved
environmentally friendly maritime power system for providing
propulsion for boats. Environmental concerns arise constantly in
the maritime sector with regard to both water and air
pollution.
[0015] A typical power system for boats consists of main propulsion
engines, propellers, donkey engines/generators, boilers, transition
and control systems etc. The main propulsion engine is the most
important component. Several types of main propulsion engines have
been developed in the marine sector including steam turbine, diesel
engines, gas turbine and nuclear engines. Among these types, diesel
engines are the most widely used and occupy .about.90% of the total
current power capacity. However, all these engines have
environmental problems. Diesel engines, steam turbines and gas
turbines need to combust fossil fuels. Contaminates (e.g. CO.sub.2,
NO.sub.x and particulates) are inevitably produced in combustion
processes. Nuclear power systems not only produce nuclear waste
pollution and provide a radiation risk but also are at least an
order of magnitude more expensive than other power systems.
[0016] Consequently, a combustion free power system with a
non-polluting exhaust would be greatly welcomed by the marine
industry and the general public. It would also be desirable if such
a marine power system could be used to generate electricity for use
within the boat and to heat and/or cool the boat as necessary.
SUMMARY OF THE INVENTION
[0017] The present invention concerns the use of a cryogenic
working fluid for energy storage, energy generation and
propulsion.
[0018] A cryogenic energy storage (CES) system according to an
embodiment of the present invention stores a cryogen produced using
electricity during off-peak hours, thus storing energy, and uses
the stored cryogen to generate electricity during peak hours, thus
releasing the stored energy. The cryogen may be pumped, heated and
then expanded in a turbine.
[0019] Accordingly, the present invention provides a method of
storing energy comprising: [0020] providing a gaseous input; [0021]
producing a cryogen from the gaseous input; [0022] storing the
cryogen; [0023] expanding the cryogen; [0024] using the expanded
cryogen to drive a turbine; and [0025] recovering cold energy from
the expansion of the cryogen.
[0026] The present invention also provides a cryogenic energy
storage system comprising: [0027] a source of cryogen; [0028] a
cryogen storage facility; [0029] means for expanding the cryogen;
[0030] a turbine capable of being driven by the expanding cryogen;
and [0031] means for recovering cold energy released during
expansion of the cryogen.
[0032] The turbine may be used to drive a generator and thus
generate electricity.
[0033] Alternatively, or in addition, the turbine may be used to
drive a propeller for example for use in a marine engine.
Consequently, the CES may be used as a Cryogenic Propulsion System
(CPS).
[0034] The turbine may comprise a multi-stage quasi-isothermal
turbine. The turbine may include reheaters or interheaters.
[0035] A number of suitable cryogens may be used. Preferably, the
cryogen comprises liquid air. Alternatively, the cryogen may
comprise slush air, liquid nitrogen, liquid hydrogen, liquid
natural gas (LNG) or any other cryogen.
[0036] The energy storage system may maximise the use of, and
minimise the modification of, current available and mature
technologies for cryogen formation, such as air liquefaction
plants.
[0037] If the cryogen comprises liquid air, the liquid air may be
produced by an air liquefaction plant and supplied to the CES at
off-peak hours. In the meantime, other products such as O.sub.2,
N.sub.2, Ar and CO.sub.2 in both gas and liquid states could be
produced as commercial products if needed. The efficiency of the
production of the cryogen may be improved by using waste cold from
other sources such as from the regasification of LNG (liquid
natural gas).
[0038] Modern large capacity cryogenic oxygen production plants
have low running costs of .about.0.4 kWh/kg (1.44 MJ/kg). This cost
is expected to decrease further to .about.0.3 kWh/kg (1.08 MJ/kg)
by 2010-2020 ("Air separation and liquefaction: recent developments
and prospects for the beginning of the new millennium", Castle W.
F., International Journal of Refrigeration, 25, 158-172, 2002;
"Energy analysis of cryogenic air separation", Cornelissen R. L.
and Hirs G. G., Energy Conservation and Management, 39, 1821-1826,
1998.). The CES may use a feedstock of liquid air from a cryogenic
plant but will work completely separately from the cryogenic plant;
this feedstock may be small depending on the `cold energy` recycle
and operation strategy. The production of liquid air may consume
about 80% of the energy required to produce liquid oxygen given
present production methods.
[0039] The cryogen may be expanded by heating. For example, the
cryogen may be heated by thermal sources including ambient,
geothermal, waste heat from power plants and/or other waste heat
resources to heat the cryogenic working fluid and generate
electricity during peak hours. The thermal sources may not
previously have been utilised for electricity generation because
the temperature difference between the working fluid and heat
source would have been considered insufficient. The working fluid
may be superheated by the waste heat. The waste heat may have
originated from power plants or from the compression process of the
input gas or even from the waste gas stream after being heated to
ambient temperature by ambient air. To increase the energy density
of the working fluid, the gaseous input may be at a high pressure
before expansion because the ideal work per unit mass of gaseous
input for an isothermal expansion for an ideal gas, W.sub.T, is
given by
W T = RT ln ( P i n P out ) ##EQU00001##
where R, T, P.sub.in, and P.sub.out are the universal constant, gas
temperature, and injection and exhaust pressures, respectively.
Moreover, the cryogen may be pumped as a liquid to a high working
pressure because little work is consumed in the pressurization of
liquid. On the other hand, the gas temperature may be as high as
possible before expansion. Use could be made of the waste heat
contained in the flue gas from power plants for heating the
cryogen. Most effectively, the ambient air could be used to heat
the cryogen to approximately the environmental temperature and the
waste heat could then be used to heat the working fluid further to
improve the energy efficiency of the entire system. Because the
temperature difference between the cryogen and ambient temperature
is high, waste heat which previously would have been considered a
poor source of energy can be used as a source of energy to heat the
cryogen.
[0040] By using the waste heat, the CES can be used as a net energy
generator. Therefore, the CES can operate as a stand-alone energy
storage plant using electricity as an energy input along with
ambient temperature heat from the atmosphere. The CES can be placed
either at the point of generation or the point of demand.
[0041] The `cold` energy contained in the cryogen as the working
fluid is very high-grade cryogenic energy and at least a portion is
recycled. In a preferred embodiment the `cold` energy contained in
the working fluid is extracted to cool down the gaseous input
(before and/or after a compressor, a fan or a blower) through heat
exchangers. The cold energy may be extracted from the exhaust gas
from the system. Assuming that the cryogen is heated to the ambient
temperature in an isobaric process before expansion, the heat
absorbed from the atmosphere by the cryogen is given by
Q=h.sub.0-h.sub.1 where h.sub.0 and h.sub.1 are enthalpy at ambient
temperature and at liquid temperature, respectively. Considering a
Carnot cycle operated between a low temperature reservoir at
T.sub.1=78.9K and a high temperature reservoir at an ambient
temperature of T.sub.0=300K, the amount of work is given by
W = Q ( T 0 T 1 - 1 ) . ##EQU00002##
Therefore the amount of work is proportional to the temperature
difference. The above equation also implies that the work required
to achieve the cold energy Q is equivalent to several multiples of
the cold energy, which should therefore be used effectively.
[0042] The input air can be compressed before, after or at the same
time as passing through the heat exchangers depending on
applications. Therefore, the compressor can be positioned either
before the heat exchanger, after the heat exchanger, or even within
the heat exchanger. If the cold air is to be used for
air-conditioning or cooling of food and other products, then it is
preferable for compression to be realised by a blower (low
pressure) located before the heat exchangers. Alternatively, if the
input air is used for producing liquid cryogen, then it is
preferable for a compressor to be placed after the heat exchangers.
Such a compressor could be a stand-alone compressor attached to the
CES if the liquifaction plant is remote to the CES. Alternatively,
if the CES is adjacent to the liquefaction plant, then the
compressor of the liquefaction plant could be used.
[0043] If the cold cryogen is used to cool the gaseous input, the
cooled gaseous input can then feed back into the cryogen plant as a
feedstock or be liquefied to cryogen inside the CES.
[0044] In addition, or alternatively, the cold energy may be used
to provide cooled air for refrigeration or air conditioning
purposes. For example, in a maritime power system, the energy
storage system can be used to drive a turbine to drive a propeller
as well as to provide cooled air for air conditioning and/or
refrigeration purposes.
[0045] Alternatively, or in addition, waste heat from the system
could be used to provide heat to the immediate environment, e.g. to
provide heating and/or hot water in a boat.
[0046] The present invention may make simultaneous use of `cold`
energy and `waste` heat. By recovering the `cold` energy from the
expansion of the stored cryogen and using it to enhance the
production of more cryogen whilst the system is operating in
electricity generation mode, the efficiency of the system as a
whole is increased. Cold energy is as useful in this system as hot
energy. In addition the CES uses energy in the ambient air (heat)
or water to heat the cryogen to close to the ambient temperature,
followed by further heating with waste heat from, for example, flue
gas and steam venting to the environment from a power generation
plant. Also, heat released from the compression of gaseous input
can also be recovered and used to heat the cryogen. The heat
applied to the cryogen causes it to expand and this drives the
cryogen.
[0047] As heat losses and hydraulic pressure drops always occur,
the pressure of the gaseous input may be increased either before or
after the one or more heat exchanger, for example at the inlet,
using, for example, a blower or a compressor. The compression
process could be adiabatic or isothermal. Assuming the ideal
behaviour of air, the work required for the isothermal process is
given by
W T = RT ln ( P 1 P 0 ) ##EQU00003##
whereas that for the adiabatic process, W.sub.Q, is given by
W Q = h 1 - h 0 = k k - 1 RT 0 [ ( P 1 P 0 ) ( k - 1 ) k - 1 ) ]
##EQU00004##
where k, P.sub.1, P.sub.0 are the specific heat ratio (=1.4 for
air), and the outlet and inlet pressures of the compressor or
blower, respectively. Therefore, the required work increases with
increasing outlet pressure P.sub.1. Therefore, P.sub.1 should be
kept as low as possible to save compression work.
[0048] Waste heat from the compressor could be used to provide heat
to the immediate environment, for example, to provide heating
and/or hot water in, for example, a boat.
[0049] In a preferred embodiment the cryogen production plant may
be integrated with the energy storage system. Alternatively, the
cryogen production plant may be remote from the energy storage
system and the cryogen could be transported between the two
plants.
[0050] A small amount of cryogen may be needed to top up the system
after each cycle.
[0051] When a non-polluting source of energy is used to power the
system, the system is environmentally benign with a potential to
reverse environmental contamination by separating environmentally
detrimental gases, such as CO.sub.2 and other contaminants,
associated with the burning of fossil fuels from the gaseous
input.
[0052] The system of the present invention does not involve any
combustion process so it will not cause any emissions. The only
working fluid is the cryogen. The effect on the environment is also
minimised because less CO.sub.2 and other environmentally
detrimental gas components such as NO.sub.X are produced or
used.
[0053] The CES system can be used for storing energy produced from
most existing power generation plants.
[0054] When the CES is configured as a CPS, the system can be used
as a propulsion device instead of in a static energy storage or
generation system. The CPS could therefore be used in a boat
engine. The CES could be configured to drive both a propeller and a
generator so that the power system could be used to both provide
propulsion and electricity for a boat.
[0055] In addition, the CPS could be further configured to provide
heat for heating a boat and/or its contents. The CPS could also be
further configured to provide cold for refrigeration purposes on
board the boat, or for air conditioning of the boat.
BRIEF DESCRIPTION OF THE DRAWINGS
[0056] The present invention will now be described in more detail
with reference to the following figures in which:
[0057] FIG. 1 shows a schematic diagram of an energy storage system
according to the present invention;
[0058] FIG. 2 shows a schematic diagram of a cryogenic air
separation and liquefaction plant;
[0059] FIG. 3 shows a schematic diagram of a CES according to the
present invention;
[0060] FIG. 4 shows a schematic diagram of a CPS according to the
present invention;
[0061] FIG. 5 shows an ideal T-S diagram of a CES according to the
present invention for an ambient pressure case;
[0062] FIG. 6 shows a practical T-S diagram of a CES according to
the present invention for an ambient pressure case;
[0063] FIG. 7 shows a practical T-S diagram of a CES with
superheating according to the present invention for an ambient
pressure case;
[0064] FIG. 8 shows a T-S diagram of a CES according to the present
invention for a low pressure ratio case;
[0065] FIG. 9 shows a T-S diagram of a CES according to the present
invention for a high pressure ratio case;
[0066] FIG. 10a shows a thermodynamic cycle for a CPS according to
the present invention;
[0067] FIG. 10b shows a thermodynamic cycle for a CPS according to
the present invention when the pressure of the input air 1 exceeds
.about.38 bar.
[0068] FIG. 11 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 0.1 MPa;
[0069] FIG. 12 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 0.2 MPa;
[0070] FIG. 13 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 0.4 MPa;
[0071] FIG. 14 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 1.0 MPa;
[0072] FIG. 15 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 2.0 MPa;
[0073] FIG. 16 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 4.0 MPa;
[0074] FIG. 17 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 10 MPa;
[0075] FIG. 18 shows four efficiencies of the thermodynamics cycles
associated with a CES according to the present invention when the
input air pressure, P.sub.1, is 20 MPa;
[0076] FIG. 19 shows the actual efficiencies of a CES according to
the present invention without superheating when the pressure of the
working fluid is 20 MPa;
[0077] FIG. 20 shows the actual efficiencies of a CES according to
the present invention with superheating when the pressure of the
working fluid is 20 MPa;
[0078] FIG. 21 shows efficiencies of a CES according to the present
invention at different turbine efficiencies when no waste heat is
used;
[0079] FIG. 22 shows efficiencies of a CES according to the present
invention at different turbine efficiencies when waste heat is
used;
[0080] FIG. 23 shows efficiencies of a CES according to the present
invention at different compressor efficiencies when no waste heat
is used;
[0081] FIG. 24 shows efficiencies of a CES according to the present
invention at different compressor efficiencies when waste heat is
used;
[0082] FIG. 25 shows efficiencies of a CES according to the present
invention at different pump efficiencies when no waste heat is
used;
[0083] FIG. 26 shows efficiencies of a CES according to the present
invention at different pump efficiencies when waste heat is
used;
[0084] FIG. 27 shows efficiencies of a CES according to the present
invention at different energy consumptions of cryogen when no waste
heat is used;
[0085] FIG. 28 shows efficiencies of a CES according to the present
invention at different energy consumptions of cryogen when waste
heat is used;
[0086] FIG. 29 shows efficiencies of a CPS according to the present
invention as a function of the pressure of input air 1;
[0087] FIG. 30 shows efficiencies of a CPS according to the present
invention as a function of the ambient temperature;
[0088] FIG. 31 shows efficiencies of a CPS according to the present
invention as a function of the efficiency of the turbine;
[0089] FIG. 32 shows efficiencies of a CPS according to the present
invention as a function of the efficiency of the compressor;
[0090] FIG. 33 shows efficiencies of a CPS according to the present
invention as a function of the efficiency of the pump;
[0091] FIG. 34 shows efficiencies of a CPS according to the present
invention as a function of the polytropic coefficients of the
compressor;
[0092] FIG. 35 shows efficiencies of a CPS according to the present
invention as a function of the isothermicity of expansion;
[0093] FIG. 36 shows efficiencies of a CES according to the present
invention as a function of temperature differences between hot and
cold fluids in the heat exchanger when no waste heat is used;
[0094] FIG. 37 shows efficiencies of a CES according to the present
invention as a function of temperature differences between hot and
cold fluids in the heat exchanger when waste heat is used;
[0095] FIG. 38 shows efficiencies of a CES according to the present
invention as a function of the temperature of the waste heat
used;
[0096] FIG. 39 shows efficiencies of a CES according to the present
invention as a function of the ambient temperature;
[0097] FIG. 40 shows efficiencies of a CPS according to the present
invention as a function of the temperature difference between hot
and cold fluids in a heat exchanger;
[0098] FIG. 41 shows efficiencies of a CPS according to the present
invention as a function of time;
[0099] FIG. 42 shows an exemplary small lab scale CES system
according to the present invention;
[0100] FIG. 43 shows a T-S diagram of the CES experimental system
of FIG. 42;
[0101] FIG. 44 shows the work output of a turbine for use in the
CES of FIG. 42 as a function of the number of stages;
[0102] FIG. 45 shows the expansion ratio of each stage of a turbine
for use in the CES of FIG. 42 as a function of the number of
stages;
[0103] FIG. 46 shows a suitable cryogenic tank for use with the CES
of FIG. 42;
[0104] FIG. 47 shows a suitable pump for use with the CES of FIG.
42;
[0105] FIG. 48 shows a suitable turbine for use with the CES of
FIG. 42;
[0106] FIG. 49 shows the characteristics of the output power and
the output duration of a number of energy storage systems;
[0107] FIG. 50 shows the relationship between the efficiency and
the cyclic period for a number of energy storage systems;
[0108] FIG. 51 shows the energy storage densities of a number of
different energy storage systems; and
[0109] FIG. 52 shows the relationship between the output power per
capital cost and the storage energy capacity per unit capital cost
for a number of different energy storage systems.
DETAILED DESCRIPTION OF THE INVENTION
[0110] A conceptual design of the energy storage system of the
present invention is shown in FIG. 1. The whole system is shown
within dotted box 100. System 100 consists of two major parts: an
air liquefaction part 200, and a Cryogenic Energy Storage unit
(CES) 300. In off-peak hours, surplus electricity is fed to the air
liquefaction plant 200 to produce liquid air, which is then used in
peak hours by the CES 300 to generate electricity. The power plant
400 and the whole energy storage system 100 only have to exchange
electricity, so no modification of the power plant 400 is needed
thus ensuring maximum flexibility. At the same time, any available
waste heat 410 from the flue gas of the power plant 400 can be used
by the CES 300 to heat the working fluid.
[0111] Within the energy storage system 100, there are two major
air streams. One stream 110 feeds air to the air liquefaction plant
200 to be liquefied and stored as liquid air in a cryogen tank.
During peak time the liquid air is pumped, heated and then expanded
in the CES 300 to generate electricity. Another air steam 120 is
input air from the atmosphere. Input air 120 is fed to the CES 300
to supply heat for expansion of the working liquid air and to
extract the `cold` energy from the working liquid air. The cooled
input air 130 can be directed to the air liquefaction plant 200 as
a feedstock or be throttled to produce liquid air within the CES
300 to reduce the amount of cryogen required from the air
liquefaction plant 200. At the same time, the air liquefaction
plant 200 can produce other products 210 such as N.sub.2, O.sub.2,
CO.sub.2, Ar etc if needed.
[0112] The cryogenic air liquefaction system 200 is a mature
technology and many types of cryogenic air liquefaction systems are
readily available "off-the-shelf". FIG. 2 shows a schematic diagram
of a typical air liquefaction plant. A liquefaction plant consists
of 5 major units: an air compression unit 220, an air pre-treatment
unit 230, an air cooling unit (not shown), a cooling unit (not
shown), and a rectification unit (not shown) (the rectification
unit is only needed if air is to be separated into different
products). The air pre-treatment unit 230 is downstream of the air
compression 220 and cooling units and is for removing contaminants
such as water, carbon dioxide, and hydrocarbons. The purified air
is then further cooled down to the cryogenic temperature using heat
exchange 240 and distilled. If needed, it is passed through the
rectification unit to produce, for example, oxygen, nitrogen, or
argon as gas or liquid products. If necessary (i.e. for air
products production), the products can be warmed up with the feed
air to conserve the refrigeration, with any deficit made up by
expanding a small portion of pressurised air.
[0113] A CES 300 according to the present invention is shown in
FIG. 3. The CES 300 comprises eight main components: compressor
310, turbine 320, generator 330, first heat exchanger 340, second
heat exchanger 350, throttling valve 360, cryogen tank 370 and pump
380.
[0114] Liquid air 250 from a cryogenic plant is introduced into the
cryogen tank 370 (in state 5 in FIG. 3) to be pumped by pump 380 to
a certain pressure (state 7). The pressurised liquid air is heated
in the second heat exchanger 350 (state 8) and then superheated in
the first heat exchanger 340 (state 9). The liquid aid, as a
working fluid, then expands to drive the turbine 320 and generator
330. The turbine 320 may be a multi-stage gas turbine with a
continuous heat supply in order to achieve a nearly isothermal
expansion. After expansion and powering of the generator 330, there
are three options for the working fluid (state 10): [0115] 1) to be
vented directly to the atmosphere and/or used for cooling or
refrigeration, [0116] 2) to be fed back into the air liquefaction
plant 200 as feedstock [0117] 3) to be introduced into the power
plant 400.
[0118] There are three possible benefits in adopting option 3:
recovery of lower grade heat, if usable, from the exhaust of the
turbine; injection into the combustion chamber of the turbine to
reduce NO.sub.x; and increasing the power output of the gas turbine
as the injected air can act as a diluent that permits greater fuel
consumption without exceeding the turbine inlet temperature limits.
These benefits may be marginal but could bring the overall
efficiency up if effectively used.
[0119] In the input air stream 120, air from the environment (state
0) is compressed (state 1) using compressor 310 and introduced to
the first heat exchanger 340 (state 2) for use in heating up the
working fluid. The compressor may be a multi-step compressor to
approach an adiabatic compression. Some unwanted components in the
input air such as water (which is bad for the turbine due to
cavitation), carbon dioxide, NO.sub.x and hydrocarbons can also be
removed during this process.
[0120] The cleaned input air then goes through the second heat
exchanger 350 (state 3) to extract more `cold energy` from the
working fluid.
[0121] The cooled input air is then either fed to the liquefaction
plant 200 as feedstock or to the throttling valve 360 to be
transformed into liquid air (state 4) for top-up of the cryogen
tank 370. A small proportion of air after the throttling is in the
gas state but is still at low temperature (state 6). This part of
cold energy is recovered by introducing the gas back into the
second heat exchanger 350. This part of the air may be rich in
oxygen so it can further be used, for example, as an oxidant in a
gas turbine or a coal-gasification turbine.
[0122] The first heat exchanger 340 may be an integrated heat
exchanger so that two parallel heat exchanging processes occur,
namely between the input air and the working fluid, and between the
working fluid and the (relatively) high temperature flue gas from
the power plant. The first heat exchanger 340 may alternatively be
designed as two separate heat exchangers, one for each of these two
processes.
[0123] FIG. 4 shows a cryogenic propulsion system (CPS) 500
according to the present invention. The CPS is based on the powered
propeller type and could offer simultaneously cold, heat,
propulsion and electricity. A CPS according to the present
invention consists of eleven major components: a propeller 505, a
turbine 510, a generator 515, a compressor 520, four heat
exchangers 525, 530, 535, 540, a throttling valve 545, a cryogen
tank 550 and a pump 555.
[0124] The working processes of the CPS system 500 comprise: [0125]
1) The liquid air from a cryogen plant or storage depot is fed into
the cryogen tank 550. [0126] 2) After being pumped, heated and
superheated, the working fluid expands to drive propeller 505
and/or generator 515 to provide propulsion and/or electricity.
[0127] 3) At the same time, an air stream from the atmosphere
(input air 1), is compressed and introduced to the heat exchangers
525, 530, 535, 540. The compression heat contained in input air 1
can be extracted via heat exchanger 525 to provide hot water/hot
air for the boat. The input air 1 then extracts the cold from the
working fluid while flowing through heat exchangers 530, 535, 540.
Finally, the input air 1 is throttled to produce liquid air and
stored in the cryogen tank 550. [0128] 4) Input air 2 and water at
the ambient temperature are introduced into heat exchanger 525 to
extract the compression heat contained in the input air 1 to
produce hot air/water as mentioned above.
[0129] Input air 3/4 under ambient conditions is introduced to
extract cold energy via heat exchangers 530 and 535 to provide cool
air for air conditioning (12.about.18.degree. C., from heat
exchanger 530) and refrigeration (-24.about.-18.degree. C., from
heat exchanger 535).
Thermodynamics Cycle Analysis--CES
[0130] Four typical cycles for the CES system of FIG. 3 are
considered in terms of the input air pressure, two at ambient
conditions, one at low pressure and one at high pressure. In the
analyses, liquid air is treated as a single phase fluid and the
gaseous air as an ideal gas. The energy losses in the compressor
310, turbine 320, pump 380 and throttle valve 360 are accounted for
by using efficiencies .eta.. For these thermodynamic analyses, the
frictional and regional losses due to flow in pipes, valves and
bends are ignored and dissipation of cryogen during storage are not
considered. The ambient temperature and pressure are expressed by
T.sub.0 and P.sub.0, respectively; the critical and boiling
temperatures of liquid air are denoted as T.sub.cr and T.sub.S,
respectively.
Ambient Input Air Pressure Case--Ideal Thermodynamics Cycle
Analysis
[0131] The ideal thermodynamics cycle is shown in FIG. 5. The
processes and the work, heat and/or exergy of these processes
are:
[0132] 1) Process 5-7, Pumping process of working fluid: The
working fluid (liquid air) from the cryogen tank is pumped from
ambient pressure P.sub.0 to P.sub.2 adiabatically. The specific
work (work per unit mass of liquid air) can be expressed by:
W 5 - 7 = V 1 ( P 2 - P 0 ) = ( P 2 - P 0 ) .rho. 1
##EQU00005##
from the viewpoint of fluid mechanics. The work can also be
expressed by the enthalpy difference between states 7 and 5 from
the first law of thermodynamics: W.sub.5-7=h.sub.7-h.sub.5.
[0133] 2) Process 7-8, Isobaric heating of working fluid: The
working fluid is heated by the input air from T.sub.s to the
ambient temperature T.sub.0. The specific work done in this process
is zero: W.sub.7-8=0. The specific heat absorbed by the working
fluid from input air is: Q.sub.7-8=h.sub.8-h.sub.7. The exergy loss
of the process is therefore:
Ex.sub.7-8=T.sub.0(S.sub.8-S.sub.7)-(h.sub.8-h.sub.7).
[0134] 3) Process 8-0, Isothermal expansion of the working fluid:
The working fluid at the high pressure expands in the turbine,
which drives the generator to generate electricity at the ambient
temperature T.sub.0. The specific ideal work done by the turbine in
this process is given by:
W.sub.8-0=T.sub.0(S.sub.0-S.sub.8)-(h.sub.0-h.sub.8). The specific
heat absorbed during the expansion by the working fluid from the
atmosphere is: Q.sub.8-0=T.sub.0(S.sub.0-S.sub.8).
[0135] 4) Process 0-6, Extraction of cold energy from the work
fluid by the input air: The input air is used to extract the cold
energy from the work fluid isobarically. No work is needed in
theory in this process: W.sub.0-6=0. The specific cold absorbed by
the input air from the working fluid is: Q.sub.0-6=h.sub.6-h.sub.0.
The exergy obtained by the input air over the process is given by:
Ex.sub.0-6=T.sub.0(S.sub.0-S.sub.6)-(h.sub.0-h.sub.6).
[0136] 5) Process 6-5, Condensation of input air: The input air is
condensed by the cold exergy released by the work fluid, which
requires zero work to be done: W.sub.6-5=0. The specific cold
energy absorbed by the input air from the work fluid is:
Q.sub.6-5=h.sub.5-h.sub.6=.lamda. where .lamda. is the latent heat
of vaporisation. The corresponding exergy obtained by the input air
is: Ex.sub.6-5=T.sub.0(S.sub.6-S.sub.5)-(h.sub.6-h.sub.5). Assuming
the mass flow of work fluid is 1, the mass flow of the input air is
x, then a heat balance gives:
Q.sub.7-8.gtoreq.x(Q.sub.0-6+Q.sub.6-5) where
Q.sub.7-8=h.sub.8-h.sub.7, Q.sub.0-6=h.sub.6-h.sub.0 and
Q.sub.6-5=h.sub.5-h.sub.6. Inserting these expressions into the
above equation gives: h.sub.8-h.sub.7.gtoreq.x(h.sub.0-h.sub.5). If
P.sub.2 is given, then h.sub.8,h.sub.7,h.sub.0,h.sub.5 can be
determined and x can be expressed by:
x .ltoreq. ( h 8 - h 7 ) ( h 0 - h 5 ) . ##EQU00006##
According to the second law of thermodynamics, exergy of a system
can only decrease without input energy, that is:
Ex.sub.7-8.ltoreq.x(Ex.sub.0-6+Ex.sub.6-5),
x .ltoreq. Ex 7 - 8 ( Ex 0 - 6 + Ex 6 - 5 ) . ##EQU00007##
Therefore, the consumption of liquid air for a single cycle is
(1-x) and the specific net work output of the cycle should be:
W.sub.net=W.sub.8-0-W.sub.5-7=T(S.sub.0-S.sub.8)-(h.sub.0-h.sub.8)-(h.sub-
.7-h.sub.5) and the energy density of CES can be expressed by:
E D = W net 1 - x = T ( S 0 - S 8 ) - ( h 0 - h 8 ) - ( h 7 - h 5 )
( 1 - x ) . ##EQU00008##
Assuming that the energy consummation of the liquid air produced in
the air liquefaction plant is E.sub.C, the energy efficiency of the
whole energy storage system (Air liquefaction+CES), E.sub.E, can be
calculated by:
E E = E D E C . ##EQU00009##
Considering the efficiency of pump .eta..sub.P and the efficiency
of turbine .eta..sub.T, the net work W.sub.net should become:
W net ' = .eta. T W 8 - 0 - 1 .eta. P W 5 - 7 = .eta. T [ T ( S 0 -
S 8 ) - ( h 0 - h 8 ) ] - ( h 7 - h 5 ) .eta. P . ##EQU00010##
The energy density of CES E.sub.D becomes:
E D ' = W net ' 1 - x = .eta. T [ T ( S 0 - S 8 ) - ( h 0 - h 8 ) ]
- ( h 7 - h 5 ) .eta. P ( 1 - x ) . ##EQU00011##
E.sub.E becomes:
E E = E D ' E C . ##EQU00012##
However, the temperature difference between the working liquid and
the input air cannot be avoided. This will decrease the temperature
T.sub.8 and increase the temperature T.sub.6. Therefore, the ideal
thermodynamics cycles overpredict the overall efficiency of the
system. This is accounted for in the following, with reference to
FIG. 6.
Ambient Input Air Pressure Case--Practical Thermodynamics Cycle
Analysis
[0137] In FIG. 6, the working liquid can only be heated to
T.sub.8', owing to the existence of a temperature difference from
the ambient temperature, and input air can only be cooled down to
T.sub.6'. Because T.sub.6' is higher than T.sub.6 (the boiling
temperature) the input air needs to be liquefied in the air
liquefaction plant, and then fed back to the CES system at state 5.
The work, heat and/or exergy related to the processes shown in FIG.
6 are given in the following:
[0138] 1) Process 5-7, Pumping process of working fluid: This
process in FIG. 6 is the same as that shown in FIG. 5. Liquid air
from the cryogen tank is pressurised by the pump from ambient
pressure P.sub.0 to P.sub.2. The specific work done on the liquid
air is:
W 5 - 7 = V 1 ( P 2 - P 0 ) = ( P 2 - P 0 ) .rho. 1
##EQU00013##
which is equal to the enthalpy difference between state 7 and state
5: W.sub.5-7=h.sub.7-h.sub.5.
[0139] 2) Process 7-8', Isobaric heating of working fluid: The
working fluid is heated by the input air from T.sub.1 to T.sub.8
instead of ambient temperature T.sub.8 (=T.sub.0). The specific
work done in this process is zero: W.sub.7-8=0. The specific heat
absorbed by the working fluid from input air is:
Q.sub.7-8'=h.sub.8'-h.sub.7. The exergy loss in the process is
therefore:
Ex.sub.7-8'=T.sub.0(S.sub.8'-S.sub.7)-(h.sub.8'-h.sub.7).
[0140] 3) Process 8'-0', Isothermal expansion of the working fluid:
The working fluid at a high pressure expands in the turbine, which
drives the generator isothermally to produce electricity. The
specific ideal work done by the turbine in this process is:
W.sub.8-0'=T.sub.0'(S.sub.0'-S.sub.8')-(h.sub.0'-h.sub.8'). The
specific heat absorbed during the expansion by the working fluid
from the atmosphere is:
Q.sub.8'-0'=T.sub.0'(S.sub.0'-S.sub.8').
[0141] 4) Process 0-6', Extraction of cold energy from the work
fluid by the input air: The input air is used to extract the cold
from the work fluid isobarically. The specific work done in this
process is zero, i.e: W.sub.0-6'=0. The specific cold extracted
from the work fluid by the input air is:
Q.sub.0-6'=h.sub.6'-h.sub.0. The exergy obtained by the input air
in the process is therefore given by:
Ex.sub.0-6'=T.sub.0(S.sub.0-S.sub.6')-(h.sub.0-h.sub.6').
[0142] 5) Process 6'-6-5, Cooling and Condensation of input air:
The input air is cooled and condensed in the air liquefaction
plant. Assuming the mass flowrate of working fluid is 1, the mass
flowrate of input air is x, heat balance of the cycle gives:
Q.sub.7-8'.gtoreq.xQ.sub.0-6' where Q.sub.7-8'=h.sub.8'-h.sub.7,
Q.sub.0-6'=h.sub.6'-h.sub.0, the above equation becomes:
h.sub.8'-h.sub.7.gtoreq.x(h.sub.0-h.sub.6'). If P.sub.2 and
temperature differences between T.sub.8 and T.sub.8' and T.sub.6
and T.sub.6' are given, h.sub.8,h.sub.7,h.sub.0,h.sub.6' can be
determined and x can then be expressed by:
x .ltoreq. ( h 8 ' - h 7 ) ( h 0 - h 6 ' ) . ##EQU00014##
According to the second law of thermodynamics, the relationship for
the exergy is: Ex.sub.7-8'.ltoreq.xEx.sub.0-6',
x .ltoreq. Ex 7 - 8 ' Ex 0 - 6 ' . ##EQU00015##
If
[0143] x = ( h 8 ' - h 7 ) ( h 0 - h 6 ' ) , ##EQU00016##
the above relation
x .ltoreq. Ex 7 - 8 ' Ex 0 - 6 ' ##EQU00017##
always holds. That implies that vaporization of 1 unit of working
fluid can pre-cool
x = ( h 8 ' - h 7 ) ( h 0 - h 6 ' ) ##EQU00018##
unit of input air. If the efficiency of heat exchanger is high
enough, then x could be greater than 1. The specific cold recycled
in this practical recycle is:
Q.sub.7-8'=xQ.sub.0-6'=x(h.sub.0-h.sub.6'). As mentioned above, the
cold energy in liquid air is very high-grade energy, assuming the
air is an ideal gas, the above cold energy is equivalent to the
ideal work given by:
W.sub.7-8'=x[T.sub.0(S.sub.0-S.sub.6')-(h.sub.0-h.sub.6')]. The
specific net work output of the cycle is therefore given by:
W net = W 8 ' - 0 ' - W 5 - 7 + W 7 - 8 ' = T 0 ' ( S 0 ' - S 8 ' )
- ( h 0 ' - h 8 ' ) - ( h 7 - h 5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h
0 - h 6 ' ) ] ##EQU00019##
and the energy density of CES is:
E D = W net 1 = T 0 ' ( S 0 ' - S 8 ' ) - ( h 0 ' - h 8 ' ) - ( h 7
- h 5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 - h 6 ' ) ] .
##EQU00020##
The energy efficiency of the whole energy storage system (air
liquefaction+CES) E.sub.E can be calculated by:
E E = E D E C . ##EQU00021##
Considering the efficiency of pump .eta..sub.P, the efficiency of
turbine .eta..sub.T and the efficiency of air liquefaction
.eta..sub.A, the net work W.sub.net should be:
W net ' = .eta. T W 8 - 0 - 1 .eta. P W 5 - 7 + W 0 - 6 ' = .eta. T
[ T 0 ' ( S 0 ' - S 8 ' ) - ( h 0 ' - h 8 ' ) ] - ( h 7 - h 5 )
.eta. P + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 - h 6 ' ) ] .
##EQU00022##
The energy density of CES E.sub.D becomes:
E D ' = W net ' 1 = .eta. T [ T 0 ' ( S 0 ' - S 8 ' ) - ( h 0 ' - h
8 ' ) ] - ( h 7 - h 5 ) .eta. P + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 -
h 6 ' ) ] ##EQU00023##
and the energy efficiency of the whole energy storage system
becomes:
E E ' = E D ' E C . ##EQU00024##
Considering further the use of waste heat, if T.sub.0 is
superheated to T.sub.9 using the waste heat from the power plant,
as shown in FIG. 7, the specific net work output of the cycle will
be:
W net 2 = W 9 - 10 - W 5 - 7 + W 7 - 8 ' = T 9 ( S 10 - S 9 ) - ( h
10 - h 9 ) - ( h 7 - h 5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 - h 6
' ) ] ##EQU00025##
and the energy density of CES is:
E D 2 = W net 2 1 = T 9 ( S 10 - S 9 ) - ( h 10 - h 9 ) - ( h 7 - h
5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 - h 6 ' ) ] .
##EQU00026##
This leads to the following energy efficiency of the entire energy
storage system (air liquefaction system+CES) E.sub.E2:
E E 2 = E D 2 E C . ##EQU00027##
If T.sub.0=300K and neglecting the energy losses due to the
turbine, pump and heat exchangers, the ideal work output for a unit
mass of liquid air can be estimated on the basis of the above
analysis by:
W net = W 8 ' - 0 ' - W 5 - 7 + W 7 - 8 ' = T 0 ' ( S 0 ' - S 8 ' )
- ( h 0 ' - h 8 ' ) - ( h 7 - h 5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h
0 - h 6 ' ) ] = 743 kJ / kg ##EQU00028##
and the ideal energy density of CES is:
E D = W net 1 = T 0 ' ( S 0 ' - S 8 ' ) - ( h 0 ' - h 8 ' ) - ( h 7
- h 5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 - h 6 ' ) ] = 180.8 kWh /
m 3 . ##EQU00029##
E.sub.C=1440 kJ/kg (0.4 kWh/kg), the ideal energy efficiency of CES
is:
If
[0144] E E = E D E C = 51.6 % ##EQU00030##
If E.sub.C=1080 kJ/kg (0.3 kWh/kg), the ideal energy efficiency of
CES becomes:
E E = E D E C = 68.8 % . ##EQU00031##
If T.sub.9 is superheated to 400K using the waste heat from the
power plant, the specific ideal work is:
W net 2 = W 9 - 10 - W 5 - 7 + W 7 - 8 ' = T 9 ( S 10 - S 9 ) - ( h
10 - h 9 ) - ( h 7 - h 5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 - h 6
' ) ] . = 881 kJ / kg ##EQU00032##
The ideal energy density of CES is:
E D 2 = W net 2 1 = T 9 ( S 10 - S 9 ) - ( h 10 - h 9 ) - ( h 7 - h
5 ) + x [ T 0 ( S 0 - S 6 ' ) - ( h 0 - h 6 ' ) ] ] = 214.3 kWh / m
3 . ##EQU00033##
If E.sub.C=1440 kJ/kg (0.4 kWh/kg), the ideal energy efficiency of
CES is:
E E = E D E C = 61.2 % . ##EQU00034##
If E.sub.C=1080 kJ/kg (0.3 kWh/kg), the ideal energy efficiency of
CES becomes:
E E = E D E C = 81.6 % . ##EQU00035##
Note that the energy consumption (0.3 and 0.4 kWh/kg) used above is
for separation of oxygen from air. The actual energy requirement of
liquid air production is approximately 80% of this figure so the
estimation of the ideal energy efficiency is conservative. On the
other hand, the probable actual efficiency is approximately 80% of
that achieved in the ideal work cycle so the efficiency as
estimated above should be close to the actual efficiency.
[0145] From the above analyses, it can be concluded that the work
output of CES increases significantly for a given amount of
cryogenic fuel consumption owing to the recovery of the cold
energy. The extra work from cold recycle is equivalent to
x[T.sub.0(S.sub.0-S.sub.6)-(h.sub.0-h.sub.6')] where x is
determined by the temperature difference and energy losses of
components. The specific work output and energy density of CES
depends on the efficiency of the turbine .eta..sub.T and the energy
consumption per unit mass of liquid air in the air liquefaction
plant E.sub.C. The efficiency of the pump is also a factor but not
as important as .eta..sub.T and E.sub.C because the work consumed
by a pump is relatively small. An increase in the temperature
differences of heat exchangers will increase the liquid air
consumption or decrease the efficiency of the cycle. It can be seen
that the energy efficiency and energy density of the energy storage
system E.sub.E is competitive to other currently available energy
systems. The system of the present invention also offers the
advantages of producing other products from the air liquefaction
plant and using the waste heat from the power plant.
Low Input Air Pressure Case--Thermodynamics Cycle of CES
Analysis
[0146] The thermodynamics cycle of a CES for a low input air
pressure case is shown in FIG. 8. Here, the term `low pressure`
denotes pressures lower than .about.3.8 MPa below which air
vaporisation is approximately isothermal. The cycle consists of the
following processes similar to those described above:
[0147] 1) Process 0-2, Isothermal pressurization of input air: The
input air is compressed isothermally from the ambient pressure
P.sub.0 to P.sub.1. The work done on the air by the compressor is:
W.sub.0-2=T.sub.0(S.sub.0-S.sub.2)-(h.sub.0-h.sub.2). The heat
Q.sub.0-2 of this isothermal process is:
Q.sub.0-2=T.sub.0(S.sub.0-S.sub.2) Unfortunately, it is difficult
to realize an absolute isothermal pressurization process, the
actual process will be a polytropic process like 0-1.
[0148] 2) Process 2-3'-3, Extraction of cold energy from the
working fluid by input air: The compressed input air is used to
extract the cold energy from the working fluid isobarically. The
work done in this process is zero: W.sub.2-3=0. The heat released
from the input air in the process 2-3 is:
Q.sub.2-3'=h.sub.3'-h.sub.2. The heat released from input air in
the process 3-3' is:
Q.sub.3'-3=h.sub.3'-h.sub.3=T.sub.3(S.sub.3'-S.sub.3)=.lamda.. The
exergy obtained from the process is therefore given by:
Ex.sub.2-3=T.sub.0(S.sub.3-S.sub.2)-(h.sub.3-h.sub.2).
[0149] 3) Process 3-4-5(-6), Throttling of compressed input air:
The compressed input air is throttled to the ambient pressure for
condensation. The work done in this process is zero: W.sub.3-4=0.
The heat released from the input air is zero: Q.sub.3-4=0.
Considering one unit of the working fluid, the total amount of
input air is x units of which a fraction y is liquefied, the amount
of liquefied air at state 5 will be xy, and the amount of gaseous
air at state 6 will be x(1-y). A heat balance over the process
3-4-5(-6) will be: h.sub.3=yh.sub.5+(1-y)h.sub.6.
[0150] 4) Process 5-7, Pumping process of working air: Process 5-7
in FIG. 8 is the same as that in FIG. 5 in which liquid air from
the cryogen tank is pumped from the ambient pressure P.sub.0 to
P.sub.2. The specific work done on the liquid air is:
W 5 - 7 = V 1 ( P 2 - P 0 ) = ( P 2 - P 0 ) .rho. 1 .
##EQU00036##
The above work can also be expressed by the enthalpy difference
between state 7 and state 5: W.sub.5-7=h.sub.7-h.sub.5.
[0151] 5) Process 7-7', Isobaric heating of the working fluid to
condense input air: The working fluid is heated to condense the
input air at T.sub.3. The specific work done in this process is
zero: W.sub.7-7'=0. The specific heat absorbed from the input air
is: Q.sub.7-7'=h.sub.7'-h.sub.7.
[0152] 6) Process 7'-8, Isobaric heating of the working fluid to
cool the input air: The working fluid is heated by the input air
from T.sub.7' to T.sub.8. The specific work done in this process is
zero: W.sub.7'-8.apprxeq.0. The specific heat absorbed from the
input air is: Q.sub.7'-8=h.sub.8-h.sub.7'. The exergy released in
the process 7-8 is:
W.sub.7-8=T.sub.0(S.sub.8-S.sub.7)-(h.sub.8-h.sub.7).
[0153] 7) Process 8-9, Isobaric superheating of the working fluid:
The working fluid is superheated by from T.sub.8 to T.sub.9 in
which no work is done, i.e.: W.sub.8-9=0 while the specific heat
absorbed from the input air over this process is:
Q.sub.8-9=h.sub.9-h.sub.8.
[0154] 8) Process 9-10, Isothermal expansion of the working fluid:
The working fluid with a high pressure expands in the turbine
isothermally which delivers work to generate electricity. The
specific ideal work done in this process is:
W.sub.9-10=T.sub.9(S.sub.10-S.sub.9)-(h.sub.10-h.sub.9). The
specific heat absorbed by the working fluid from the ambient in the
process is: Q.sub.9-10=T.sub.9(S.sub.10-S.sub.9). It should be
noted that the T.sub.9 is higher than the ambient temperature,
which requires energy from the waste heat from the power plant to
ensure an isothermal expansion. If the expansion of air is an
adiabatic process, the specific ideal work W.sub.ad will be:
W ad = k k - 1 RT 9 [ P 0 P 2 ) ( k - 1 ) k - 1 ] ##EQU00037##
which means no heat absorption namely: Q.sub.ad=0. The actual work,
however, is expected to be in the range between W.sub.9-10 and
W.sub.ad. A factor called isothermicity .gamma. is often used as an
index, which is defined as the ratio of the actual work to the
isothermal work:
.gamma. = W ac W 9 - 10 . ##EQU00038##
Thus, the actual work W.sub.ac can expressed as:
W ac = .gamma. W 9 - 10 = .gamma. RT 9 ln ( P 2 P 0 ) .
##EQU00039##
[0155] 9) Process 6-6', Extraction of cold from exhaust air to
condense the input air: The exhaust air (part of input air after
the throttling) is used to condense the input air isobarically. The
specific work done in this process is zero: W.sub.6-6'=0. The
specific heat absorbed from input air is:
Q.sub.6-6'=h.sub.6'-h.sub.6. The heat balance of 3'-3, 7-7' and
6-6' is therefore given by:
xQ.sub.3-3'=Q.sub.7-7'+x(1-y)Q.sub.6-6',
x(h.sub.3-h.sub.3')=(h.sub.7'-h.sub.7)+x(1-y)(h.sub.6'-h.sub.6).
[0156] 10) Process 6'-0, Extraction of cold from exhaust air to
cool the input air: The exhaust air is used to cool down the input
air isobarically. The specific work done in this process is zero:
W.sub.6-40 -0=0. The specific cold absorbed from the exhaust air by
the input air is: Q.sub.6'-0=h.sub.0-h.sub.6'. The heat balance of
2-3', 7'-8 and 6'-0 is expressed as:
xQ.sub.2-3'.ltoreq.Q.sub.7'-8+x(1-y)Q.sub.6'-0,
x(h.sub.2-h.sub.3').ltoreq.(h.sub.8-h.sub.7')+x(1-y)(h.sub.0-h.sub.6').
The exergy obtained in process 6-0 is:
Ex.sub.0-6=T.sub.0(S.sub.0-S.sub.6)-(h.sub.0-h.sub.6). From the
heat and exergy balances of the cycle, x and y can be calculated by
the following equations based on the T-S diagram in FIG. 8):
{ h 3 = yh 5 + ( 1 - y ) h 6 x ( h 3 - h 3 ' ) = ( h 7 ' - h 7 ) +
x ( 1 - y ) ( h 6 ' - h 6 ) x ( h 2 - h 3 ' ) .ltoreq. ( h 8 - h 7
' ) + x ( 1 - y ) ( h 0 - h 6 ' ) xEx 2 - 3 .ltoreq. Ex 7 - 8 + x (
1 - y ) Ex 0 - 6 . ##EQU00040##
From the above equations, the ratio of liquefaction of the input
air y is:
y = ( h 6 - h 4 ) ( h 6 - h 5 ) . ##EQU00041##
Because (h.sub.6-h.sub.5)>(h.sub.6-h.sub.4)>0 always holds,
therefore 1>y>0. Similarly, x can be expressed as:
x = ( h 7 ' - h 7 ) ( h 3 - h 3 ' ) - ( 1 - y ) ( h 6 ' - h 6 ) .
##EQU00042##
As
(h.sub.7-h.sub.7)>0,[(h.sub.3-h.sub.3')-(1-y)(h.sub.6'-h.sub.6)]>-
;0 always holds, x>0. This means that vaporisation of one unit
of the working fluid can produce xy units of liquid air, and the
consumption of the cycle will be (1-xy). As a consequence, the
specific net work output of the cycle is:
W net = W 9 - 10 - W 5 - 7 - xW 0 - 2 = [ T 9 ( S 10 - S 9 ) - ( h
10 - h 9 ) ] - ( h 7 - h 5 ) - x ( T 0 ( S 0 - S 2 ) )
##EQU00043##
and the energy density of CES can be expressed by:
E D = W net 1 - xy = [ T 9 ( S 10 - S 9 ) - ( h 10 - h 9 ) ] - ( h
7 - h 5 ) - x ( T 0 ( S 0 - S 2 ) ) 1 - xy . ##EQU00044##
The energy efficiency of the entire energy storage system (Air
liquefaction system+CES) E.sub.E can therefore be calculated
by:
E E = E D E C . ##EQU00045##
Considering the efficiencies of the pump .eta..sub.P, the turbine
.eta..sub.T and the compressor .eta..sub.COM, the net work
W.sub.net should be:
W net ' = .eta. T W 9 - 10 - W 5 - 7 .eta. P - W 0 - 2 .eta. COM =
.eta. T [ T 9 ( S 10 - S 9 ) - ( h 10 - h 9 ) ] - ( h 7 - h 5 )
.eta. P - x ( T 0 ( S 0 - S 2 ) ) .eta. COM , ##EQU00046##
the energy density of CES E.sub.D becomes:
E D ' = W net ' 1 - xy = .eta. T [ T 9 ( S 10 - S 9 ) - ( h 10 - h
9 ) ] - ( h 7 - h 5 ) .eta. P - x ( T 0 ( S 0 - S 2 ) ) .eta. COM 1
- xy ##EQU00047##
and E.sub.E becomes:
E E = E D ' E C . ##EQU00048##
[0157] Based on the above analysis, it can be concluded that, in
comparison with cryogenic (liquid nitrogen) powered engines, the
consumption of cryogenic fuel is reduced by xy for 1 unit of
working fluid but with a penalty of work required for compression
W.sub.0-2=T.sub.0(S.sub.0-S.sub.2). The specific work output will
be improved as the penalty is less than the benefit due to the
reduction in the working fluid consumption. As the work of
compression is much less than the work output of the turbine, the
specific work output and the energy density of CES mainly depend on
the efficiency of the turbine CT and energy consumption for air
liquefaction. This is similar to the case of using the ambient
pressure. The efficiencies of pump and compressor are not key
factors for improving the work output and energy density of CES.
The efficiency of this cycle is expected to be lower than that of
FIG. 6 because the process of isothermal condensation has low
energy efficiency.
High Input Air Pressure Case--Thermodynamics Cycle of CES
Analysis
[0158] The thermodynamic cycle of the CES for a high input air
pressure case is shown in FIG. 9. Here, the term `high input air
pressure` means the pressure is higher than 3.8 MPa above which air
has no isothermal vaporisation process. The processes of this case
are as follows:
[0159] 1) Process 0-2, Isothermal pressurization of input air: The
input air is compressed from ambient pressure P.sub.0 to P.sub.1
isothermally. The work done on the air by the compressor is:
W.sub.0-2=T.sub.0(S.sub.0-S.sub.2)-(h.sub.0-h.sub.2). The heat
Q.sub.0-2 of this isothermal process is:
Q.sub.0-2=T.sub.0(S.sub.0-S.sub.2) Unfortunately, it is difficult
to realize an absolute isothermal pressurisation process, the
actual process will be 0-1.
[0160] 2) Process 2-3, Extraction of cold energy from the working
air by input air: The compressed input air is used to extract the
cold energy from the work fluid isobarically. The work done in this
process is zero: W.sub.2-3=0. The heat released from the input air
for process 2-3 is: Q.sub.2-3=h.sub.3-h.sub.2. The exergy obtained
from the process is:
Ex.sub.2-3=T.sub.0(S.sub.3-S.sub.2)-(h.sub.3-h.sub.2).
[0161] 3) Process 3-4-5(-6), Throttling of compressed input air:
The compressed input air is throttled to the ambient pressure for
condensation. The work done in this process is zero: W.sub.3-4=0.
The heat released from the input air is zero: Q.sub.3-4=0. Similar
to the low input air pressure case, considering one unit of working
fluid and assuming a total of x units of input air of which a
fraction y is liquefied, the amount of liquid air produced by
liquefaction at state 5 is xy, and the amount of gaseous air at
state 6 is x(1-y). The heat balance of 3-4-5(6) is therefore
expressed as: h.sub.3=yh.sub.5+(1-y)h.sub.6.
[0162] 4) Process 5-7, Pumping of working fluid: This process is
the same as that in FIG. 6. Liquid air from the cryogen tank is
pumped from ambient pressure P.sub.0 to P.sub.2. The work done on a
unit mass of liquid air is:
W 5 - 7 = V l ( P 2 - P 0 ) = ( P 2 - P 0 ) .rho. l .
##EQU00049##
The work can also be expressed by enthalpy difference between the
states 7 and 5: W.sub.5-7=h.sub.7-h.sub.5.
[0163] 5) Process 7-8, Isobaric heating of the working fluid to
cool input air: The working fluid is heated to condense the input
air at T.sub.3, and there is work involved in this process:
W.sub.7-8=0. The specific heat absorbed from the input air is:
Q.sub.7-8=h.sub.8-h.sub.7. The exergy released in the process 7-8
is therefore:
W.sub.7-8=T.sub.0(S.sub.8-S.sub.7)-(h.sub.8-h.sub.7).
[0164] 6) Process 8-9, Isobaric superheating of the working fluid:
The working fluid is superheated by the input air from T.sub.8 to
T.sub.9 in which zero work is done, i.e. W.sub.8-9=0. The specific
heat absorbed from input air is: Q.sub.8-9=h.sub.9-h.sub.8.
[0165] 7) Process 9-10: Isothermal expansion of the working fluid:
The working fluid with a high pressure expands in the turbine and
delivers work isothermally. The specific ideal work done in this
process is: W.sub.9-10=T.sub.9(S.sub.10-S.sub.9)-(h.sub.10-h.sub.9)
while the specific heat absorbed in the process is:
Q.sub.9-10=T.sub.9(S.sub.10-S.sub.9). Similar to the low pressure
case, T.sub.9 is higher than the ambient temperature; the waste
heat from the power plant is needed to keep this process
isothermal. If the expansion of the working fluid is adiabatic, the
specific ideal work W.sub.ad will be:
W ad = k k - 1 RT 9 [ P 0 P 2 ) ( k - 1 ) k - 1 ] .
##EQU00050##
The specific heat absorbed in the process is: Q.sub.ad=0. As a
result of the above analysis, the actual work should be in the
range between W.sub.9-10 and W.sub.ad. As mentioned before, the
isothermicity .gamma. is used to describe the non-ideality:
.gamma. = W ac W 9 - 10 . ##EQU00051##
Thus, the actual work W.sub.ac should be expressed as:
W.sub.ac=.gamma.W.sub.9-10=.gamma.[T.sub.9(S.sub.10-S.sub.9)-(h.sub.10-h.-
sub.9)].
[0166] 8) Process 6-0, Extraction of cold energy from the exhaust
air to cool the input air: The exhaust air after the throttling is
used to cool the input air isobarically. The specific work done in
this process is zero: W.sub.6-0=0. The specific cold absorbed by
the input air is: Q.sub.6-0=h.sub.0-h.sub.6. The heat balance over
processes 2-3, 7-8 and 6-0 is expressed as:
xQ.sub.2-3=Q.sub.7-8+x(1-y)Q.sub.6-0,
x(h.sub.2-h.sub.3)=(h.sub.8-h.sub.7)+x(1-y)(h.sub.0-h.sub.6). The
exergy obtained in the process 6-0 is:
Ex.sub.0-6=T.sub.0(S.sub.0-S.sub.6)-(h.sub.0-h.sub.6). Based on the
heat and exergy balances of processes 2-3, 3-4-5-6, 7-8, 6-0, x and
y can be calculated by the following equations on the basis of a
T-S diagram for the air:
{ h 3 = yh 5 + ( 1 - y ) h 6 x ( h 2 - h 3 ) = ( h 8 - h 7 ) + x (
1 - y ) ( h 0 - h 6 ) xEx 2 - 3 .ltoreq. Ex 7 - 8 + x ( 1 - y ) Ex
0 - 6 . ##EQU00052##
From the above equations, the ratio of liquefaction of the input
air y can be expressed by:
y = ( h 6 - h 4 ) ( h 6 - h 5 ) . ##EQU00053##
Similar to the method in the low pressure case, 1>y>0 always
holds, and x can be expressed as:
x = ( h 8 - h 7 ) ( h 2 - h 3 ) - ( 1 - y ) ( h 0 - h 6 ) .
##EQU00054##
As
(h.sub.8-h.sub.7)>0,[(h.sub.2-h.sub.3)-(1-y)(h.sub.0-h.sub.6)]>0
always holds, one has x>0. This means vaporisation of one unit
of working fluid could produce xy units of liquid air, while the
consumption of this cycle is 1-xy, and the specific net work output
of the cycle will be:
W net = W 9 - 10 - W 5 - 7 - xW 0 - 2 = [ T 9 ( S 10 - S 9 ) - ( h
10 - h 9 ) ] - ( h 7 - h 5 ) - x ( T 0 ( S 0 - S 2 ) )
##EQU00055##
and the energy density of CES is:
E D = W net 1 - xy = [ T 9 ( S 10 - S 9 ) - ( h 10 - h 9 ) ] - ( h
7 - h 5 ) - x ( T 0 ( S 0 - S 2 ) 1 - xy . ##EQU00056##
The energy efficiency of the entire energy storage system (Air
liquefaction+CES) E.sub.E can therefore be calculated by:
E E = E D E C . ##EQU00057##
Considering the efficiencies of the pump .eta..sub.P, the turbine
.eta..sub.T and the compressor .eta..sub.COM, one has the following
net work W.sub.net:
W net ' = .eta. T W 9 - 10 - W 5 - 7 .eta. P - W 0 - 2 .eta. COM =
.eta. T [ T 9 ( S 10 - S 9 ) - ( h 10 - h 9 ) ] - ( h 7 - h 5 )
.eta. P - x ( T 0 ( S 0 - S 2 ) .eta. COM . ##EQU00058##
As a result of the above, the energy density of CES E.sub.D
becomes:
E D ' = W net ' 1 - xy = .eta. T [ T 9 ( S 10 - S 9 ) - ( h 10 - h
9 ) ] - ( h 7 - h 5 ) .eta. P - x ( T 0 ( S 0 - S 2 ) .eta. COM 1 -
xy ##EQU00059##
and E.sub.E becomes:
E E = E D ' E C . ##EQU00060##
From the above analysis, it can be seen that, compared with the
design of liquid nitrogen powered engines, the consumption of
cryogenic fuel for this cycle is decreased by xy but with a penalty
of work by W.sub.0-2=T.sub.0(S.sub.0-S.sub.2)-(h.sub.0-h.sub.2).
However, the specific work output is improved due to the decrease
of liquid fuel consumption. The work required by the compressor
should be comparable with that produced by the turbine. As a
consequence, the efficiency of the compressor .eta..sub.COM becomes
a key parameter determining the overall efficiency of the CES. This
cycle is more suitable for producing liquid air through the CES
part of the energy storage system.
[0167] The above thermodynamics analyses on the four typical cycles
show that: [0168] 1) The energy efficiency and energy density of
CES are improved in comparison with liquid nitrogen powered engines
owing to the cold energy recycle. [0169] 2) The overall performance
of the energy storage system is determined by the efficiency of the
turbine, and the specific work output and the specific energy
consumption of the air liquefaction plant. [0170] 3) The
temperature differences across the heat exchangers will increase
the liquid air consumption thus decreasing the efficiency of the
cycle. [0171] 4) The energy efficiency and density of the CES will
be improved if the waste heat from the power plant is utilised.
[0172] The results also show that the efficiency of the CES is
competitive to other energy storage systems. Additionally, the
system can make use of the waste heat and produce air products if
needed.
Thermodynamics Cycle Analysis--CPS
[0173] FIG. 10a shows thermodynamic cycles for a CPS according to
the present invention. There are four air steams which are denoted
by the following lines: working fluid--line 580; input air 1--line
585; input air 2--line 590; and input air 3--line 595. In the
analyses, liquid air is treated as a single phase fluid and the
gaseous air as an ideal gas. The energy losses in the compressor
520, turbine 510, and pump 555 are accounted for by using their
efficiencies .eta.. For these thermodynamic analyses, the
frictional and regional losses due to flow in pipes, valves and
bends are ignored and dissipation of cryogen during storage are not
considered. The ambient temperature and pressure are expressed by
T.sub.0 and P.sub.0, respectively; the boiling temperature of
liquid air is denoted as T.sub.S.
[0174] 1) 1-2: Pumping process of working fluid: The working fluid
(liquid air) from the cryogen tank is pumped from the ambient
pressure P.sub.0 to P.sub.2. The specific work done on the liquid
air is:
W 1 - 2 = V l ( P 2 - P 0 ) = ( P 2 - P 0 ) .rho. l .
##EQU00061##
The above work can also be expressed by the enthalpy difference
between state 2 and state 1: W.sub.1-2=h.sub.2-h.sub.1.
[0175] 2) 2-2': Isobaric heating of working fluid to condense input
air 1: The working fluid is heated to condense the input air at
T.sub.7. The specific work done in this process is zero:
W.sub.2-2'=0. The specific heat absorbed from the input air 1 is:
Q.sub.2-2'=h.sub.2'-h.sub.2.
[0176] 3) 2'-3: Isobaric heating of working fluid: The working
fluid is heated by the input air from T.sub.2 to T.sub.3. The
specific work done in this process is zero: W.sub.2'-3=0. The
specific heat absorbed from the input air is:
Q.sub.2'-3=h.sub.3-h.sub.2'. The exergy released in the process 2-3
is: Ex.sub.2-3=T.sub.0(S.sub.3-S.sub.2)-(h.sub.3-h.sub.2).
[0177] 4) 3-0: Isothermal expansion of working fluid. The working
fluid with a high pressure expands in the turbine isothermally
which delivers work to generate propulsion and electricity. The
specific ideal work done in this process is:
W.sub.3-0=T.sub.0(S.sub.0-S.sub.3)-(h.sub.0-h.sub.3). The specific
heat absorbed from the ambient in the process is:
Q.sub.3-0=T.sub.0(S.sub.0-S.sub.3). If the expansion of the working
fluid is adiabatic, the specific ideal work W.sub.ad will be:
W ad = k k - 1 RT 0 [ P 0 P 2 ) ( k - 1 ) k - 1 ] ##EQU00062##
which means no heat absorption, namely: Q.sub.ad=0. The actual
work, however, is expected to be in the range between W.sub.3-0 and
W.sub.ad. A factor called isothermicity, .gamma., is often used as
an index, which is defined as the ratio of the actual work to the
isothermal work:
.gamma. = W ac W 3 - 0 . ##EQU00063##
Thus, the actual work W.sub.ac can be expressed as:
W.sub.ac=.gamma.W.sub.3-0=.gamma.[T.sub.0(S.sub.0-S.sub.3)-(h.sub.0-h.sub-
.3)].
[0178] 5) 0-4: Polytropic pressurisation of input air 1: The input
air 1 is compressed polytropically from the ambient pressure
P.sub.0 to P.sub.1. The work done on the air by the compressor
is:
W 0 - 4 = n n - 1 RT 0 [ ( P 1 P 0 ) ( n - 1 ) n - 1 ]
##EQU00064##
where n is the polytropic coefficient. The heat, Q.sub.0-4, of this
polytropic process is: Q.sub.0-4=c.sub.n(T.sub.4-T.sub.0) where
C.sub.n is the polytropic heat ratio:
c n = n - k n - 1 C V . ##EQU00065##
T.sub.4 can be calculated by:
T 4 T 0 = ( P 1 P 0 ) n - 1 n . ##EQU00066##
[0179] 6) 4-5: Release of heat from input air 1 isobarically to
input air 2: The heat of the input air 1 is released to the input
air 2 or water to produce hot air/water. The work done in this
process is zero: W.sub.4-5=0. The heat released from the input air
1 in the process 4-5 is: Q.sub.4-5=h.sub.4-h.sub.5.
[0180] 7) 5-6-7: Cooling of input air 1 by working fluid: The
compressed input air 1 is cooled by working fluid isobarically and
the cold energy inside the working fluid is extracted at the same
time. The work done in this process is zero: W.sub.5-7=0. The heat
released from the input air 1 in the process 5-6 is:
Q.sub.5-6=h.sub.5-h.sub.6. The heat released from input air in the
process 6-7 is:
Q.sub.6-7=h.sub.6-h.sub.7=T.sub.6(S.sub.6-S.sub.7)=.lamda.. The
exergy obtained from the process is therefore given by:
Ex.sub.5-6-7=T.sub.0(S.sub.5-S.sub.7)-(h.sub.5-h.sub.7).
[0181] 8) 7-8-9(-1): Throttling of compressed input air 1: The
compressed input air 1 is throttled to the ambient pressure for
condensation. The work done in this process is zero: W.sub.7-8=0.
The heat released from the input air is zero: Q.sub.7-8=0.
[0182] 9) 9-9': Extraction of cold from exhaust air to condense the
input air: The exhaust air (part of input air 1 after the
throttling) is used to condense the input air isobarically. The
specific work done in this process is zero: W.sub.9-9'=0. The
specific heat absorbed from input air is:
Q.sub.9-9'=h.sub.9'-h.sub.9.
[0183] 10) 9'-0: Extraction of cold from exhaust air to cool the
input air: The exhaust air (part of the input air 1) is used to
cool down the input air 1 isobarically. The specific work done in
this process is zero: W.sub.9'-0=0. The specific cold absorbed from
the exhaust air by the input air 1 is: Q.sub.9-40
-0=h.sub.0-h.sub.9'. The exergy obtained in process 9-0 is:
Ex.sub.9-0=T.sub.0(S.sub.0-S.sub.9)-(h.sub.0-h.sub.9).
[0184] 11) 0-10: Extraction of cold energy of the working fluid by
input air 3 isobarically for air conditioning: The cold of the
working fluid is extracted by input air 3 for cool air production
to be used for air conditioning. The work done in this process is
zero: W.sub.0-10=0. The cold energy from the working fluid in the
process 0-10 is: Q.sub.0-10=h.sub.0-h.sub.10.
[0185] 12) 0-10-11: Extraction of cold energy of the working fluid
by input air 4 isobarically for refrigeration: The cold of the
working fluid is extracted by input air 4 for refrigeration. The
work done in this process is zero: W.sub.0-11=0. The cold energy
from the working fluid in the process 0-11 is:
Q.sub.0-11=h.sub.0-h.sub.11.
[0186] 13) 0-12: Extraction of heat energy of the input air 1 by
input air 2/water isobarically: The heat of the input air 1 is
extracted by input air 2/water for hot air/water production. The
work done in this process is zero: W.sub.0-12=0. The heat released
from the input air 1 in the process 0-12 is:
Q.sub.0-12=h.sub.12-h.sub.0.
Analysis of Energy Balance
[0187] Assuming that, for one unit of the working fluid, the total
amount of input air 1 is x.sub.1, the total amount of input air 2
is x.sub.2, the total amount of input air 3/4 is x.sub.3+x.sub.4
with x.sub.3 units for air condition and x.sub.4 units for
refrigeration. In the one unit of the working fluid, a.sub.1 units
are used for the input air 1, a.sub.2+a.sub.3 units for cooling
input air 3/4 in which a.sub.2 is for x.sub.3 and a.sub.3 is for
x.sub.4. According to the first and second laws of thermodynamics,
the following balances of heat and exergy can be obtained:
[0188] 1) Heat balance in process 7-8-9(-1): Assuming that, for
a.sub.1 units of the working fluid, y fraction of the input air 1
is liquefied, the amount of liquefied air at State 1 will be
x.sub.1y, and the amount of gaseous air at State 9 will be
x.sub.1(1-y). A heat balance over the process 7-8-9(-1) will be:
h.sub.7=yh.sub.1+(1-y)h.sub.9. From the above equations, ratio of
liquefaction of the input air y is:
y = ( h 9 - h 7 ) ( h 9 - h 1 ) . ##EQU00067##
Because (h.sub.9-h.sub.1)>(h.sub.9-h.sub.7)>0 always holds,
therefore 1>y>0.
[0189] 2) Heat balance in processes 6-7, 2-2' and 9-9': The heat
balance of 6-7, 2-2' and 9-9' can be given by:
x.sub.1Q.sub.6-7=a.sub.1Q.sub.2-2'+x.sub.1(1-y)Q.sub.9-9',
x.sub.1(h.sub.6-h.sub.7)=a.sub.1(h.sub.2'-h.sub.2)+x.sub.1(1-y)(h.sub.9'--
h.sub.9). x.sub.1 can be expressed as:
x 1 = a 1 ( h 2 ' - h 2 ) ( h 6 - h 7 ) - ( 1 - y ) ( h 9 ' - h 9 )
. ##EQU00068##
[0190] 3) Heat balance in processes 5-6, 2'-3 and 9'-0: The heat
balance of 5-6, 2'-3 and 9'-0 can is expressed as:
x.sub.1Q.sub.5-6.ltoreq.a.sub.1Q.sub.2-3+x.sub.1(1-y)Q.sub.9'-0,
x.sub.1(h.sub.5-h.sub.6).ltoreq.a.sub.1(h.sub.3-h.sub.2)+x(1-y)(h.sub.0-h-
.sub.9').
[0191] 4) Heat balance in processes 4-5 and 0-12: The heat balance
in processes 4-5 and 0-12 can be expressed by: x.sub.2
(h.sub.12-h.sub.0)=x.sub.1(h.sub.5-h.sub.4). x.sub.2 can be
expressed as:
x 2 = x 1 ( h 5 - h 4 ) ( h 12 - h 0 ) . ##EQU00069##
[0192] 5) Heat balance in processes 0-10 and 2-3: The heat balance
in processes 0-10 and 2-3 can be expressed by:
x.sub.3(h.sub.0-h.sub.10)=a.sub.2 (h.sub.3-h.sub.2). x.sub.3 can be
expressed as:
x 3 = a 2 ( h 3 - h 2 ) ( h 0 - h 10 ) . ##EQU00070##
[0193] 6) Heat balance in processes 0-11 and 2-3: The heat balance
in processes 0-11 and 2-3 can be expressed by:
x.sub.4(h.sub.0-h.sub.11)=a.sub.3(h.sub.3-h.sub.2). x.sub.4 can be
expressed as:
x 4 = a 3 ( h 3 - h 2 ) ( h 0 - h 11 ) . ##EQU00071##
[0194] 7) Exergy balance of processes 5-7, 2-3 and 9-0: The exergy
balance of processes 5-7, 2-3 and 9-0 can be given by:
x.sub.1Ex.sub.5-7.ltoreq.a.sub.1Ex.sub.2-3+x.sub.1(1-y)Ex.sub.0-9,
x.sub.1[T.sub.0(S.sub.5-S.sub.7)-(h.sub.5-h.sub.7)].ltoreq.a.sub.1[T.sub.-
0(S.sub.3-S.sub.2)-(h.sub.3-h.sub.2)]+x.sub.1[T.sub.0(S.sub.0-S.sub.9)-(h.-
sub.0-h.sub.9)]
8) Mass Conservation of Process 2-3:
[0195] a.sub.1+a.sub.2+a.sub.3=1
Analysis of the Efficiency and Energy Density
[0196] The following analyses use the efficiency of work
(electricity) defined as:
.eta. w = W output W input ##EQU00072##
where W.sub.output and W.sub.input are the total works converted by
the input and output energies, respectively. To calculate the
equivalent work of heat and cold energy, two coefficients of
performance (COP), refrigeration COP (.epsilon.), and heat pump COP
(.zeta.), are used in the conversion of heat to work. As a
consequence, the specific net work output of the cycle
W output = W 3 - 0 - W 1 - 2 - x 1 W 0 - 4 + 1 .zeta. x 2 Q 0 - 12
+ 1 1 x 3 Q 0 - 10 + 1 2 x 4 Q 0 - 11 ##EQU00073##
is:
= [ T 0 ( S 0 - S 3 ) - ( h 0 - h 3 ) ] - ( h 2 - h 1 ) - x 1 n n -
1 RT 0 [ P 0 P 1 ) ( n - 1 ) n - 1 ] + 1 .zeta. x 2 ( h 12 - h 0 )
+ 1 1 x 3 ( h 0 - h 10 ) + 1 2 x 4 ( h 0 - h 11 ) .
##EQU00074##
On the other hand,
[0197] the productivity of liquid air by input air 1 is x.sub.1y.
Therefore the consumption of the working fluid is (1-x.sub.1y). The
energy density of CPS can be expressed by:
E D = W output 1 - x 1 y . ##EQU00075##
It is known that the maximum specific work of liquid air, W.sub.R,
is: W.sub.R=T.sub.0(S.sub.0-S.sub.1)-(h.sub.0-h.sub.1). The energy
efficiency of the CPS, E.sub.E, can therefore be calculated by:
E E = W output ( 1 - x 1 y ) W R . ##EQU00076##
Considering the efficiencies of the pump .eta..sub.P, the turbine
.eta..sub.T and the compressor .eta..sub.COM, the net work
W.sub.output should be:
W output ' = .eta. T W 3 - 0 - W 1 - 2 .eta. p - x 1 W 0 - 4 .eta.
COM + 1 .zeta. x 2 Q 0 - 12 + 1 1 x 3 Q 0 - 10 + 1 2 x 4 Q 0 - 11 =
1 .eta. T [ T 0 ( S 0 - S 3 ) - ( h 0 - h 3 ) ] - 1 .eta. P ( h 2 -
h 1 ) - x 1 .eta. COM n n - 1 RT 0 [ P 0 P 1 ) ( n - 1 ) n - 1 ] ,
+ 1 .zeta. x 2 ( h 12 - h 0 ) + 1 1 x 3 ( h 0 - h 10 ) + 1 2 x 4 (
h 0 - h 11 ) ##EQU00077##
the energy density of CPS, E.sub.D, becomes:
E D ' = W net ' 1 - x 1 y ##EQU00078##
and E.sub.E becomes:
E E ' = W output ' ( 1 - x 1 y ) W R . ##EQU00079##
[0198] Based on the above analysis, it can be concluded that:
[0199] 1) The maximum specific work W.sub.R gives the upper limit
of the energy density of CPS. If ambient temperature T.sub.0=300K
is used, the value is .about.743 kJ/kg. [0200] 2) If there is no
cold energy recycling, the ideal specific work output will be
W.sub.output=W.sub.3-0-W.sub.1-2. The practical value
[0200] W output ' = .eta. T W 3 - 0 - W 1 - 2 .eta. p
##EQU00080##
gives the lower limit of the CPS. If ambient temperature
T.sub.0=300K is used and the working pressure of liquid air is 200
bar, assuming the efficiencies turbine and pump are both 0.78, the
specific work output would be .about.326 kJ/kg.
[0201] If the pressure of the input air 1 exceeds .about.38 bar,
there will be no isothermal condensing process in FIG. 10a. The T-S
diagram of this case is shown in FIG. 10b. The thermodynamic
analysis is similar to the case in FIG. 10a.
Parametric Analysis--CES
[0202] A computational code has been written in the Fortran 90
environment to simulate the influences of various parameters on the
performance of the CES system. The code is written for
thermodynamics cycles operated between pressures above the ambient
pressure and 3.8 MPa (see FIG. 8), which is the most complicated
case. The code can be used easily for high pressure cases (see FIG.
9) and the ambient condition (see FIGS. 5 to 7). Six parameters
have been considered including: [0203] Pressure of the working
fluid (P.sub.2), [0204] Pressure of the input air (P.sub.1), [0205]
Efficiency of the turbine (.eta..sub.T), [0206] Efficiency of the
compressor (.eta..sub.COM), [0207] Efficiency of the pump
(.eta..sub.P), [0208] Efficiency of the air liquefaction plant
(.eta..sub.A).
[0209] The effects of these six parameters on four efficiencies
related to the performance of the CES have been analysed. The four
efficiencies that have been considered are: [0210] the efficiency
of the ideal cycle without superheating (E.sub.E), [0211]
efficiency of the ideal cycle with superheating (E.sub.Sup), [0212]
efficiency of the practical cycle without superheating (E'.sub.E)
[0213] efficiency of the practical cycle with superheating
(E'.sub.Sup).
Pressure of the Working Fluid (P.sub.2)
[0214] The four efficiencies of the thermodynamics cycles
associated with the CES are shown in FIGS. 11 to 18 under nine
different pressures of the working fluid (P.sub.2=0.2 MPa, 0.4 MPa,
1.0 MPa, 2.0 MPa, 4.0 MPa, 10 MPa, 20 MPa, 30 MPa, 40 MPa and 50
MPa) at different input air pressures (P.sub.1). The ambient
temperature is assumed to be T.sub.0=300K, the superheat
temperature is taken as T.sub.9=400K, and the efficiencies of the
turbine, compressor and pump are assumed as 0.88
(.eta..sub.T=.eta..sub.COM=.eta..sub.P=0.88). The temperature
differences of the heat exchangers are not considered at this
stage. This will be discussed below.
[0215] At P.sub.1=0.1 MPa (FIG. 11), which represents a
thermodynamic cycle at the ambient pressure (FIGS. 5 to 7), all
four of the efficiencies increase with increasing pressure of the
working fluid (P.sub.2). However, the increase is only significant
at pressures of P.sub.2<.about.10 MPa above which the curves
level off. At pressures of P.sub.2>20 MPa the efficiencies are
almost constant. The maximum efficiencies are found to be
E.sub.E=0.507, E.sub.Sup=0.640, E'.sub.E=0.459 and
E'.sub.Sup=0.569, respectively.
[0216] At pressures of P.sub.1=0.2-2.0 MPa (FIGS. 12 to 15), which
represent thermodynamics cycles at low pressure (FIG. 8), the
results are similar to the case at the ambient pressure (see FIG.
11). That is, all four of the efficiencies increase sharply with
increasing P.sub.2 until P.sub.2 reaches 10 MPa when farther
increase in the efficiencies is very small. A comparison between
FIG. 11 and FIGS. 12 to 15 reveals that the efficiencies at
P.sub.1=0.2-2.0 MPa are lower than those at P.sub.1=0.1 MPa.
[0217] At P.sub.1=4.0-20 MPa (FIGS. 16-18), which represent
thermodynamics cycles at high pressures (FIG. 9), the efficiencies
of the practical cycle without superheating (E'.sub.E) are
significantly lower than those for P.sub.1<2.0 MPa owing to the
consumption of compression of the input air. The efficiencies of
the practical cycle with superheating (E'.sub.Sup) are high because
the heat from the superheating is treated as a waste and a large
proportion of the liquid air can be produced by the CES.
[0218] From the above analysis, it can be concluded that P.sub.2
should be higher than 10 MPa. However, the selection of P.sub.2 may
be limited by the mechanical feasibility. At present,
pressurisation of air to 20 MPa is very common practice in the air
separation and liquefaction plants without any engineering
difficulties. According to the analysis, P.sub.2=20 MPa is
recommended for the CES as pressures higher than 20 MPa lead to a
very marginal increase in efficiency. As a consequence, the
following analyses are all based on P.sub.2=20 MPa.
Pressure of the Input Air (P.sub.1)
[0219] The actual efficiencies of CES without and with superheating
are plotted in FIGS. 19 and 20 respectively as a function of
pressure of the input air (P.sub.1) for the given pressure of the
working fluid (P.sub.2=20 MPa). Three efficiencies of the turbine,
compressor and pump (.eta..sub.T=.eta..sub.P=.eta..sub.COM=0.80,
0.84, 0.88) are considered. From inspection of these figures it can
be seen that the actual efficiencies increase with increasing
efficiencies of the three components (turbine, compressor and pump)
with and without superheating. Without superheating, the maximum
efficiency occurs at the ambient pressure (P.sub.1=0.1 MPa) and the
efficiency (E'.sub.E) decreases sharply with increasing input air
pressure. With superheating, the efficiency decreases sharply first
with increasing input air pressure (P.sub.1) between 0.1 and 0.4
MPa. A further increase of P.sub.1 between 0.4 and .about.2 MPa
leads to little change in the efficiency. However, a further
increase in P.sub.1 to .about.4 MPa results in a large increase in
the efficiency due to production of a large proportion of liquid
air. A further increase in P.sub.1 beyond 4 MPa leads to a decrease
in the efficiency due to increasing compression work. There are two
peaks in the efficiency plots with the peak values depending on the
efficiency of three components (turbine, compressor and pump). For
an efficiency of the components of 0.88, the best efficiency of CES
occurs at P.sub.1=4 MPa. For an efficiency of the components of
0.80 and 0.84, the best CES efficiency occurs at P.sub.1=0.1
MPa.
[0220] Therefore, if no waste heat is used by the CES system,
P.sub.1=0.1 MPa should be selected as the working pressure of the
input air, as the efficiency is highest and there is no need for a
compressor, hence reducing the capital investment and maintenance
costs. As a result of this analysis, the following analyses are
conducted under the two pressure conditions of P.sub.1=0.1 MPa and
P.sub.1=4.0 MPa.
Efficiency of Turbine (.eta..sub.T)
[0221] As mentioned above, two sets of conditions are considered,
namely, (P.sub.1=0.1 MPa, P.sub.2=20 MPa), and (P.sub.1=4.0 MPa,
P.sub.2=20 MPa). The efficiencies of the compressor and pump are
taken as 0.88 (.eta..sub.COM=.eta..sub.P=0.88). The ambient
temperature is assumed as T.sub.0=300K, the superheat temperature
is T.sub.9=400K. The temperature differences across the heat
exchangers are not considered. Simulations are performed with seven
turbine efficiencies of 0.68, 0.72, 0.76, 0.80, 0.88, 0.92, 0.96
and 1.00 and the results are illustrated in FIGS. 21 and 22,
without and with heat recycle respectively. The efficiencies of CES
for both cases increase monotonically with increasing efficiency of
the turbine. However, the dependence of the efficiency of the CES
is a function of P.sub.1, the turbine efficiency and the use of
waste heat. An increase in the efficiency of the turbine by one
percent leads to an increase in the CES efficiency by 0.318% for
P.sub.1=0.1 MPa without heat recycle, an increase of 0.690% for
P.sub.1=0.1 MPa with heat recycle, an increase of 0.428% for
P.sub.1=4.0 MPa without the heat recycle, and an increase of by
2.742% for P.sub.1=4.0 MPa with heat recycle.
[0222] The figures also show that the rate of increase in the CES
efficiency at P.sub.1=0.1 MPa is lower than that at P.sub.1=4.0
MPa, indicating that the cycle efficiency at P.sub.1=4.0 MPa relies
more on the efficiency of the turbine than does the cycle
efficiency at P.sub.1=0.1 MPa.
[0223] If there is no waste heat, the CES efficiency at P.sub.1=0.1
MPa is higher than that at P.sub.1=4.0 MPa for a turbine efficiency
of from 0.68 to 1.0. This indicates that P.sub.1=0.1 MPa should be
used for the CES operation in the absence of the waste heat
recycle.
[0224] If waste heat is used, the CES efficiency at P.sub.1=0.1 MPa
is lower than that at P.sub.1=4.0 MPa for a turbine efficiency over
0.80, but the reverse is seen when the turbine efficiency is lower
than .about.0.8. As a consequence, there is a need for
optimisation.
Efficiency of Compressor (.eta..sub.COM)
[0225] The effect of the compressor efficiency on the CES
efficiency is illustrated in FIGS. 23 and 24, without and with heat
recycle respectively. Simulations are carried out for seven
compressor efficiencies of 0.68, 0.72, 0.76, 0.80, 0.88, 0.92, 0.96
and 1.00 with the following conditions: P.sub.1=0.1 or 4.0 MPa,
P.sub.2=20 MPa, T.sub.0=300K, T9=400K, and
.eta..sub.T=.eta..sub.P=0.88. The temperature differences across
the heat exchangers are not considered.
[0226] The efficiency of the CES cycle for P.sub.1=0.1 MPa and
P.sub.2=20 MPa is constant as no compression of the input air is
needed for P.sub.1=0.1 MPa. The efficiency of the CES at
P.sub.1=4.0 MPa and P.sub.2=20 MPa increase monotonically with
increasing efficiency of the compressor. An increase in the
compressor efficiency by one percent leads to an increase in the
CES efficiency by 0.717% for P.sub.1=4.0 MPa without heat recycle,
and an increase in the CES efficiency by 1.056% for P.sub.1=4.0 MPa
with heat recycle. This indicates that for P.sub.1=4.0 MPa, the
efficiency of the compressor contributes significantly to the CES
efficiency.
[0227] From FIG. 23, one can see that the efficiency of the CES at
P.sub.1=0.1 MPa is much higher than that at P.sub.1=4.0 MPa when
there is no waste heat recycle. If the waste heat is available,
then FIG. 24 shows that the efficiency of the CES cycle at
P.sub.1=0.1 MPa is lower than that at P.sub.1=4.0 MPa if the
efficiency of the compressor is higher than 0.78, and the reverse
is seen for compressor efficiencies lower than 0.78.
Efficiency of Pump (.eta..sub.P)
[0228] Simulations are performed on seven pump efficiencies of
0.68, 0.72, 0.76, 0.80, 0.88, 0.92, 0.96 and 1.00 for P.sub.1=0.1
or 4.0 MPa, P.sub.2=20 MPa, T.sub.0=300K, T.sub.9=400K and
.eta..sub.T=.eta..sub.COM=0.88. The temperature differences across
the heat exchangers are not considered. The results are illustrated
in FIGS. 25 and 26, without and with heat recycle respectively,
from which one can see that the efficiencies of both CES cycles
increase monotonically with increasing pump efficiency. However,
the increase is very small; an increase in the pump efficiency by
one percent only leads to an increase in the efficiency of the CES
cycle by 0.025% for P.sub.1=0.1 MPa without heat recycle, by 0.068%
for P.sub.1=0.1 MPa with heat recycle, by 0.022% for P.sub.1=4.0
MPa without heat recycle, and by 0.072% for P.sub.1=4.0 MPa with
heat recycle.
[0229] This indicates that the efficiency of the CES depends little
on the efficiency of the pump because the work consumed by the pump
is about an order of magnitude smaller than that of turbine and the
compressor.
Efficiency of Air Separation Plant (.eta..sub.A)
[0230] FIGS. 27 and 28 show the efficiencies of the CES as a
function of energy consumption per kilogram of liquid air produced.
Six levels of energy consumption of 0.400, 0.375, 0.350, 0.325,
0.300 and 0.275 kWh/kg are considered, which correspond
respectively to an efficiency of the air separation plant of
.eta..sub.A=0.516, 0.559, 0.602, 0.645, 0.688 and 0.731. The
rationale for these levels of energy consumption is that the
current energy consumption of liquid air production is .about.0.4
kWh/kg, and it is expected to decrease to .about.0.28-0.3 kWh/kg by
2010.about.2020. Other conditions are P.sub.1=0.1 or 4.0 MPa,
P.sub.2=20 MPa, T.sub.0=300K, T.sub.9=400K and
.eta..sub.7=.eta..sub.P=.eta..sub.COM=0.88.
[0231] The results show that the efficiency of the CES increases
monotonically with a decrease in the energy consumption of cryogen
production. An increase in the efficiency of the air separation
plant by one percent results in an increase in the efficiency of
the CES cycle by .about.0.972% for P.sub.1=0.1 MPa without heat
recycle, an increase in the efficiency of the CES cycle by
.about.1.181% for P.sub.1=0.1 MPa with heat recycle, an increase in
the efficiency of the CES cycle by 0.590% for P.sub.1=4.0 MPa
without heat recycle, and an increase in the efficiency of the CES
cycle by 1.381% for P.sub.1=4.0 MPa with heat recycle.
[0232] Compared with efficiencies of the turbine, compressor and
pump, the efficiency of the air liquefaction plant is a more
important factor contributing significantly to the overall
efficiency of the CES.
[0233] If the energy consumption of the liquid air production were
reduced to .about.0.28 kWh/kg, then the efficiency of the CES
without waste heat recycle would be increased to .about.0.670 and
that with waste heat recycle to .about.0.951.
[0234] Accordingly, the results of the above parametric analysis
show that P.sub.1=0. MPa and P.sub.2=20.0 MPa give the best
performance for cases without waste heat recycle. The results also
show that P.sub.1=4.0 MPa and P.sub.2=20.0 MPa could give a better
performance for cases with waste heat recycle than P.sub.1=0.1 MPa
and P.sub.2=20.0 MPa could do, depending on the efficiencies of the
components of the CES. The efficiencies of the turbine
(.eta..sub.T), the compressor (.eta..sub.COM), and the air
separation plant (.eta..sub.A) are shown to be the most important
parameters in determining the overall CES efficiency, whereas the
pump efficiency (.eta..sub.P) has very little influence on the
performance of CES.
[0235] Two parts of energy have been incorporated into the total
possible energy from liquid air: a) isothermal expansion of
compressed gas to ambient pressure and b) cold exergy utilisation
by pre-cooling the air input for the separation and liquefaction
system. For a simple idealised case (P.sub.1=0.1 MPa, P.sub.2=20
MPa, T=300 K), the ideal work from liquid air could be .about.740
kJ/kg that includes the contribution of a) 450 kJ/kg and b) 290
kJ/kg.
[0236] For the gas expansion work (450 kJ/kg) non-isothermal
expansion is inevitable. An external heat source has to be added to
maintain a high isothermity. A conventional turbine may achieve
energy efficiency up to 85% under optimised conditions. It is
visible that similar efficiency could be achieved for the proposed
turbine applications. However, due to the very high pressures
.about.200 bar involved, multi-stage expansion could be considered.
The nearly-ambient operation temperature of the turbine also
requires considerations of the sealing and lubrication issues.
[0237] For the recycling of the cold exergy (290 kJ/kg) the amount
of the cold exergy that can be recycled is dependent on the a)
operational pressures model, b) charging and discharging modes, and
c) existence of extra cold energy storage system.
[0238] For the operational pressure models, two optimum cases have
been identified: I) input air at 0.1 MPa and the working fluid at
20.0 MPa for operation temperature .about.300 k (no waste heat
added) and II) input air at 0.1 MPa or 4.0 MPa and working fluids
at 20.0 MPa for operation temperature .about.400 K (with a waste
heat recycle). Take the optimized case I) for example (P1=0.1 Mpa,
P2=20 Mpa and T=300K), for an ideal compression (dS=0), the
temperature of liquid air after compressing to 20 MPa is .about.84
K, which is the lowest temperature that incoming air can get. The
liquefaction process needs to remove .about.230 kJ/kg (sensible
heat) and another 200 kJ/kg (latent heat) at saturation
temperature, e.g. 78 K for air under 1 bar. The only work that can
be saved through a heat exchanger is some of the work needed to
reduce temperature from ambient to .about.84 K (this normally
involves multi-stage compression and throttling for a liquefaction
factory). Approx. 50% of the energy (latent heat+some sensible
heat) for air liquefaction can not be extracted by the cold energy
from the heat exchanger. The extra electricity needed is .about.0.2
kWh/kg air during discharging hours if a 0.4 kWh/kg industrial rate
is assumed.
[0239] For the optimized case II) where the waste heat is utilized,
the analysis will be the same as above for the incoming air at 0.1
MPa (P1=0.1 MPa, P2==20 MPa and T=400K), but different for the
incoming air pressure at 4 MPa (P.sub.1=4 MPa, P2=20 MPa and
T=400K).
[0240] Since the saturation temperature is .about.131K at 4 Mpa,
incoming air can be cooled down directly to the liquid state
through a heat exchanger, which means no extra electricity is
needed for manufacturing liquid air during peak hours. However this
comes with the penalty of the compression work needed to bring air
to 4 Mpa. For a pure isothermal compression, .about.0.328 MJ/kg,
.about.0.1 kWh electricity, is needed. The temperature rise is also
significant: for an adiabatic compression with compressor
efficiency of 0.9, the temperature rise is .about.620K. The
temperature rise reduces to 283 K and 132K respectively for a poly
constant of 1.2 and 1.1. Extra cooling facilities are requested for
the compressor to achieve a nearly-isothermal compression.
[0241] The amount of cold energy recycled is also dependent on the
flow rate ratio during charging and discharging periods. The cold
exergy application (in the energy release process) is based on the
simultaneous cooling of incoming air (in the energy storage
process) in a liquefaction unit. In principle these two events do
not occur at the same time. For a typical energy storage system,
the duration of the energy release process is only a couple of
hours in peak times. To maintain safety and extend the running
time, a typical liquefaction unit will operate fall load at
off-peak time and continue to operate at low load at other times.
For a model with 8 hrs of discharging and 16 hrs of charging, the
steady flow ratio is .about.2:1. If running at a 50% load during
peak times, the flow rate ratio is increased to 4:1. For every
kilogram of liquid air produced at peak times, only the sensible
heat .about.230 kJ/kg-air can be cooled down by the evaporation of
liquid air for the optimized operational pressure case I).
Therefore, a large amount of cold energy will not be fully
utilized. The shorter the discharging ratio, the larger the amount
of cold energy wasted. To fully utilize the energy, the load for
liquefaction could be increased but this would risk the consumption
of more electricity at peak times.
[0242] Alternatively, the cold energy could be stored. During the
discharging period, part of the cold energy could be used to
pre-cool the incoming air. At the same time, the extra part of the
cold energy will be stored in a thermal energy storage system (TES)
that will release cold during the off-peak time to pre-cool the
incoming air. This could maximize the opportunities of using cold
energy. The storage material may include phase change materials,
cryogenic storage materials and others. The storage material is
chosen based on its thermal conductivity, specific heat, thermal
diffusivity, density, and kinetic behaviour etc. The rate of heat
absorption and releasing is directly related to the energy
efficiency especially for the phase change materials. The energy
storage system may be in the form of fixed bed, suitable geological
sites and others. The storage efficiency may be influenced by the
properties of the storage materials, the storage temperature and
pressure, and the heat transfer coefficient between gas and storage
materials.
Parametric Analysis--CPS
[0243] A computational code has been written in the Fortran 90
environment to simulate the influences of various parameters on the
performance of the CPS system. The code is written for
thermodynamics cycles operated between pressures above the ambient
pressure and 38 bar (see FIG. 10a), which can also be used for the
high pressure case (see FIG. 10b). Seven parameters have been
considered including: [0244] Pressure of the input air 1 (P.sub.1),
[0245] Ambient temperature (T.sub.0), [0246] Efficiency of the
turbine (.eta..sub.T), [0247] Efficiency of the compressor
(.eta..sub.COM), [0248] Efficiency of the pump (.eta..sub.P),
[0249] Polytropic coefficient of compression (n), [0250]
Non-isothermicity of expansion in turbine.
[0251] In the simulations, the pressure of working fluid is taken
as 200 bar, the temperature of hot air/water supplied by CPS as 328
K (55.degree. C.), the temperature of cold air for air condition
supplied by CPS as 285 K (12.degree. C.), the temperature of cold
air for food refrigeration supplied by CPS as 249 K (-24.degree.
C.), the coefficient of performance (COP) of the heat pump (.zeta.)
as 3.0, the COP of cooling air for air conditioning
(.epsilon..sub.1) as 5.0, and the COP of refrigeration
(.epsilon..sub.2) as 3.0.
Pressure of Input Air 1 (P.sub.1)
[0252] The ideal and actual efficiencies of CPS are plotted in FIG.
29 under 14 different pressures of the input air 1 (P.sub.1=1.0
bar, 2.0 bar, 3.0 bar, 4.0 bar, 6.0 bar, 8.0 bar, 10 bar, 12 bar,
14 bar, 16 bar, 18 bar, 20 bar, 30 bar, 40 bar). The ambient
temperature is taken as 300K, the polytropic coefficient of the
compressor is taken as 1.2 and three efficiencies of the turbine,
compressor and pump are considered
(.eta..sub.T=.eta..sub.P=.eta..sub.COM=0.88, 0.84, 0.80).
[0253] From inspection of FIG. 29, it can be seen that the
efficiency of CPS increases with increasing efficiencies of the
three components (turbine, compressor and pump). The efficiency
increases first with increasing pressure of input air 1 and then
decreases after reaching a peak. The maximum efficiency is found to
be 0.793, 0.679, 0.646, 0.613 when .eta..sub.COM=1 (ideal), 0.88,
0.84 and 0.80, respectively. For an ideal case, the peak efficiency
of the CPS occurs at P.sub.1=.about.14 bar. The optimal pressure of
input air 1 at which the peak occurs decreases with decreasing
efficiency of the components, namely, P.sub.1=8 bar for 0.88,
.about.6 bar for 0.84 and 0.80.
[0254] A high pressure of the input air 1 can produce a high
proportion of liquid air and therefore a further increase in the
efficiency of CPS. However, a high pressure of the input air 1 also
consumes more compression work. Therefore, an optimal pressure of
the input air 1 should be selected for the best CPS performance.
Since the optimal pressure is not significantly different for the
three realistic efficiencies of 0.88, 0.84 and 0.80, the pressure
of the input air 1 is selected as 8 bar, and the following
calculations are based on this pressure. At P.sub.1=8 bar,
P.sub.2=200 bar, .eta..sub.T=.eta..sub.P=.eta..sub.COM=0.88, the
maximum energy efficiency of CPS is 67.7%, and the specific outputs
of the work, heat and cold of the CPS are 401.9 kJ/kg, 29.4 kJ/kg,
342.8 kJ/kg, respectively. It can be seen that the amount of cold
produced by CPS is very large. Therefore, the CPS is particularly
suitable for refrigeration boats.
Ambient Temperature (T.sub.0)
[0255] FIG. 30 shows the influence of the ambient temperature on
the efficiency of the CPS at P.sub.1=8 bar and P.sub.2=200 bar for
five ambient temperatures of 270K, 280K, 290K, 300K and 310K with
n=1.2 and .eta..sub.T=.eta..sub.P=.eta..sub.COM=0.88. When the
ambient temperature is 270K, 280K or 290K, it is considered
unnecessary to account for cool air for air conditioning. It is
apparent that the efficiency of CPS increases monotonically with
increasing ambient temperature. When the ambient temperature
increases from 270K to 310K, the efficiency of the CPS is increased
by 14.9%. Due to the utilisation of cold energy for air
conditioning at temperatures higher than 290K, there is a sharp
increase in the efficiency from 290K to 300K, It is therefore
concluded that the CPS performs better in locations with a high
ambient temperature such as tropical regions.
Efficiency of Turbine (.eta..sub.T)
[0256] FIG. 31 shows the influence of the efficiency of the turbine
on the overall efficiency of the CPS for seven values of
.eta..sub.T=0.68, 0.72, 0.76, 0.80, 0.88, 0.92, 0.96 and 1.00 with
.eta..sub.COM=.eta..sub.P=0.88, n=1.2, T.sub.0=300K, P.sub.2=200
bar, and P.sub.1=8 bar. The efficiency of CPS increases almost
linearly with increasing efficiency of the turbine. An increase in
the efficiency of the turbine by 1% leads to an increase in the CPS
efficiency by 0.738%. The efficiency of the turbine is therefore a
key parameter for the CPS efficiency.
Efficiency of Compressor (.eta..sub.COM)
[0257] The effect of the compressor efficiency on the CPS
efficiency is illustrated in FIG. 32. Simulations are carried out
for seven compressor efficiencies of 0.68, 0.72, 0.76, 0.80, 0.88,
0.92, 0.96 and 1.00 with P.sub.1=8 bar, P.sub.2=200 bar,
T.sub.0=300K, .eta..sub.T=.eta..sub.P=0.88 and n=1.2. The
efficiency of the CPS increases monotonically with increasing
efficiency of the compressor. An increase in the compressor
efficiency by 1% leads to an increase in the CPS efficiency by
0.09%. Therefore the efficiency of the compressor does not
contribute significantly to the CPS efficiency. This is because the
amount of work consumed by the compressor is small due to the
relatively low working pressure of input air 1 compared to that of
the working fluid, and the relatively low flow rate of input air 1
compared to that of the working fluid due to a considerable part of
the cold energy of the working fluid being used to provide cold air
for air conditioning and refrigeration.
Efficiency of Pump (.eta..sub.P)
[0258] The effect of the pump efficiency on the CPS efficiency is
illustrated in FIG. 33. Simulations are performed on seven pump
efficiencies of 0.68, 0.72, 0.76, 0.80, 0.88, 0.92, 0.96 and 1.00
for P.sub.1=1 bar, P.sub.2=200 bar, T.sub.0=300K,
.eta..sub.T=.eta..sub.COM=0.88 and n=1.2. The efficiency of the CPS
increases monotonically with increasing pump efficiency. However,
the rate of increase is very small; an increase in the pump
efficiency by 1% only leads to an increase in the CPS efficiency by
0.0625%. Therefore the efficiency of the CPS depends little on the
efficiency of the pump.
Polytropic Coefficient of Compression (n)
[0259] Simulations are performed on seven polytropic coefficients
of 1.05, 1.10, 1.15, 1.20, 1.25, 1.30, and 1.35 for P.sub.1=1 bar,
P.sub.2=200 bar, T.sub.0=300K,
.eta..sub.T=.eta..sub.COM=.eta..sub.P=0.88. The results are shown
in FIG. 34, from which it can be seen that the efficiency of the
CPS changes little with increasing polytropic coefficient. This is
because the amount of work consumed by the compressor is small, and
the compression heat is recycled by the input air 2.
Isothermicity of Expansion (.gamma.)
[0260] Five values of the isothermicity of the expansion process in
the turbine of 0.80, 0.85, 0.90, 0.95, 1.0 for P.sub.1=1 bar,
P.sub.2=200 bar, T.sub.0=300K,
.eta..sub.T=.eta..sub.COM=.eta..sub.P=0.88 are simulated and the
results are shown in FIG. 35. The efficiency of the CPS increases
almost linearly with increasing isothermicity. An increase in the
isothermicity by 1% gives an increase in the CPS efficiency by
0.72%. The isothermicity of the expansion is therefore a key
parameter for the CPS efficiency.
Heat Transfer Analysis--CES
[0261] The heat exchangers are critical components of the CES. Heat
exchangers are widely used in the cryogenics and air liquefaction
industries, which has led to establishment of a substantial
technology base. In general, the following factors are considered
when designing a heat exchanger:
(1) Heat transfer requirement (2) Efficiency or temperature
differences of the exchanger (3) The dimensions of the space
available (4) The need for low heat capacity (5) The cost (6) The
importance of pressure drop (7) The operating pressure
[0262] In the following analysis, focus is on the assessment of
heat transfer requirements, type and size of the exchangers and
influences of various factors on the performance heat exchangers.
The following assumptions are made:
(1) Thermodynamic equilibrium between fluid phases (2) Even flow
distribution within heat exchanger (3) Fully developed turbulent
flow (4) Adiabatic shell wall (5) Zero axial conduction (6) No
radiation between hot and cold fluids (7) Constant overall heat
transfer coefficient.
[0263] Considering a heat exchanger that exchanges an amount of
heat, Q, between a hot and a cold fluid, the volume of the heat
exchangers can be assessed by:
V = S = Q q = Q U _ .DELTA. T _ ##EQU00081##
where V represents the volume of the heat exchanger, S is the heat
transfer area, .theta. is the ratio of the compactness of heat
exchangers defined as
= S V , ##EQU00082##
q is the heat flux, is the overall average heat transfer
coefficient, and .DELTA. T is the average temperature difference
between the hot and cold fluids. Considering a tube-in-shell
configuration, the overall heat transfer coefficient may be
obtained by:
1 U _ = 1 U i + 1 U w + 1 U o ##EQU00083##
where U.sub.i is the heat transfer coefficient between the tube
wall and the tube side fluid, U.sub.o is that between the tube wall
and the shell side fluid, and U.sub.w accounts for the heat
conductivity across the tube wall expressed by:
U w = .lamda. .delta. ##EQU00084##
with .lamda.,.delta. respectively the wall thermal conductivity and
wall thickness. There is a large amount of literature on the
calculations of the heat transfer coefficients U.sub.i and U.sub.o.
For a turbulent flow in a smooth cylindrical tube, the heat
transfer coefficient between the tube side fluid and the tube wall
is given approximately by Nu=0.023Re.sup.0.8Pr.sup.0.4 where Nu is
the Nusselt number defined as
N u = UD .lamda. , ##EQU00085##
Re is the Reynolds number defined as
Re = .rho. uD .mu. , ##EQU00086##
and Pr is the Prandtl number given by
Pr = v .alpha. , ##EQU00087##
where .rho. is the density of the fluid, D is the diameter of the
tube, .nu. is the fluid kinematic viscosity, .alpha. is the fluid
thermal diffusivity and .mu. is the fluid dynamic viscosity. For
the pressure drop of a Newtonian fluid in a smooth cylindrical
tube, the frictional pressure drop can be calculated by
.DELTA. p = 2 f .rho. u 2 L D ##EQU00088##
where f is the friction factor, .mu. is the flow velocity, L is the
length of the tube. For a turbulent flow in tubes, the Blasius
equation is generally used for estimating f in a wide range of
Reynolds number:
f = 0.079 Re 0.25 . ##EQU00089##
The flow in the heat exchangers in the CES of the present invention
is likely to be in the two-phase region for which a full analysis
of the pressure drop requires a 3-dimensional description of the
flow and heat transfer involving phase changes. An engineering
approach is to calculate first the pressure with a homogeneous
model and then use a safety factor of 3.about.5 in the design of
the heat exchangers.
Heat Transfer Requirement
[0264] The CES system could have up to four heat exchangers:
(1) Heat exchanger 1 (350): for input air to extract cold from the
working fluid (and take heat from the ambient air) (2) Heat
exchanger 2 (340): for waste heat to superheat the working fluid
(3) Heat exchanger 3: for the turbine to absorb heat from
atmosphere (4) Heat exchanger 4: for the compressor to ensure
isothermal operation.
[0265] The specific heat transfer requirements for the four heat
exchangers are, respectively:
Heat Exchanger 1 Q.sub.1=h.sub.8-h.sub.7 Heat Exchanger 2
Q.sub.2=h.sub.9-h.sub.8
Heat Exchanger 3 Q.sub.3=T.sub.9(S.sub.10-S.sub.9)
Heat Exchanger 4 Q.sub.4=T.sub.0(S.sub.0-S.sub.2)
[0266] By using the above four equations, the specific heat
transfer requirements under different conditions are obtained and
illustrated in Table 1, where the ambient temperature is 300 K and
the superheat temperature is 400 K.
[0267] From table 1 it can be seen that for P.sub.1=0.1 MPa, since
there is no need for a compressor, Q4 is zero. If there is no
superheat, Q2 is zero. The total quantity of specific heat transfer
requirement for a simple cycle without superheating at P.sub.1=0.1
MPa is therefore 858.6 kJ/kg. The maximum specific heat transfer
requirement is 1308.2 kJ/kg, which corresponds to the cycle with
superheating at P.sub.1=4.0 MPa. In the following sections,
analyses will be based on the above two sets of heat transfer
requirements.
TABLE-US-00001 TABLE 1 Specific heat transfer requirement at
different conditions P2 P1 Q.sub.1 Q.sub.2 Q.sub.3 Q.sub.4 (MPa)
(MPa) Super Heat (kJ/kg) (kJ/kg) (kJ/kg) (kJ/kg) 20.0 0.1 No 368.9
0.0 489.7 0.0 20.0 0.1 Yes 368.9 119.8 630.8 0.0 20.0 4.0 No 310.3
0.0 489.7 247.3 20.0 4.0 Yes 310.3 119.8 630.8 247.3
Preliminary Design of Heat Exchanger
[0268] As mentioned above, there is a substantial technology base
for heat exchangers and there are a lot of types of heat exchangers
available for the cryogenics and air separation and liquefaction
industries. Tube-in-shell and plate-and-fin heat exchangers are
among the most widely used types. Tube-in-shell heat exchangers are
commonly used at relatively high temperatures. Tube-in-shell heat
exchangers have a high transfer coefficient ranging from .about.300
to .about.3000W/m.sup.2K when the fluid phase in both the shell and
tube sides is liquid. A common technique to improve the performance
of tube-in-shell heat exchangers is to foil fins helically around
the tube thus forming a tube and fin heat exchanger in order to
increase the ratio of compactness and the heat transfer
coefficient. This is especially effective when the fluid is in a
gaseous state in one or both sides of the heat exchanger. In
addition, the temperature difference between hot and cold fluids is
relative high (.about.15 K), which leads to a relatively low
efficiency.
[0269] Plate and fin heat exchangers have an advantage of a high
degree of compactness, and a low temperature difference between the
hot and cold fluids. This type of heat exchanger can be made of
aluminium alloy so the capital cost is relatively low. Plate and
fin heat exchangers are suitable for use in the cryogenic field
because the innate flexibility of this type of heat exchanger
allows the use of a multiplicity of fluids in the same unit. Plate
and fin heat exchangers comprise flat plates of aluminium alloy
separated by corrugated fins. The fins are brazed onto the plate by
means of a thin foil of the same alloy as the plate with added
silicon to cause melting of the foil at low temperatures and so to
bond the fins to the plate. Aluminium is generally favoured on
grounds of cost but copper is also acceptable. Due to the large
ratio of compactness of .about.250.about.5000 m.sup.2/m.sup.3,
plate and fin heat exchangers are the most widely used heat
exchangers in the air separation and liquefaction industry with a
typical heat transfer coefficient of .about.30.about.500 W/m.sup.2K
and a temperature difference of up to .about.2.about.6 K between
the hot and cold fluids.
[0270] Other types of heat exchangers that could be used include
regenerators, coiled tube heat exchangers, multiple tube heat
exchangers, and coaxial tube heat exchangers.
[0271] The following is an estimation of the size of the heat
exchangers based on the performance of the plate and fin type. The
overall average heat transfer coefficient is taken as 100
W/m.sup.2K; the average temperature difference between hot and cold
fluids, .DELTA. T, is assumed as 2 K; the ratio of compactness,
.theta., is taken as 1000 m.sup.2/m.sup.3. The compactness could be
much higher, so the estimation is on the conservative side. The
maximum heat transfer requirements, H.sub.R, with and without
superheating are respectively given as 858.6 and 1308.2 kJ/kg on
the basis of the above calculation. Two cases of the CES with
electricity storage volumes (Ev) of 1 MWh and 500 MWh are
considered in the estimation. The operating time (O.sub.T) of the
CES is assumed as 8 hours. This is according to peak hour
operation. Different duty cycles could be used and should not
greatly impact efficiency.
[0272] For Case 1 of the CES with the storage volume of 1 MWh, the
heat transfer requirement without superheating is given by:
Q = E V H R O T E D = 0.149 MW ##EQU00090##
where E.sub.D is the energy density of liquid air (kJ/kg). The
total size of the heat transfer exchangers can be calculated
by:
V = Q U _ .DELTA. T _ = 0.745 m 3 . ##EQU00091##
[0273] For Case 1 with the superheating, the heat transfer
requirement will be:
Q = E V H R O T E D = 0.186 MW . ##EQU00092##
The total size of the heat transfer exchangers will be:
V = Q U _ .DELTA. T _ = 0.929 m 3 . ##EQU00093##
[0274] For Case 2 of the CES with a storage volume of 500 MWh, the
heat transfer requirement without superheating is:
Q = E V H R O T E D = 74.5 MW . ##EQU00094##
The total size of the heat exchangers will be:
V = Q U _ .DELTA. T _ = 372.5 m 3 . ##EQU00095##
If the heat exchanger is assumed to be cubic in shape, the length
of each side will be 7.19 m. If a factor of safety of 4 is given,
the length of each side will be 11.41 m.
[0275] For Case 2 of the CES with the storage volume of 500 MWh
with superheating, the heat transfer requirement will be:
Q = E V H R O T E D = 93.0 MW . ##EQU00096##
The total size of the heat exchangers will be:
V = Q U _ .DELTA. T _ = 464.5 m 3 . ##EQU00097##
If a cubic shape is assumed, then the length of each side of the
heat exchanger will be 7.74 m. If a factor of safety of 4 is given,
the length of each side will be 12.29 m.
[0276] The liquid nitrogen viscous pressure drop is reported to be
about 0.05 MPa and the pressure drop of input air is about 400 Pa.
If a safety factor of 4 is used, then the liquid air pressure drop
would be about 0.2 MPa which is about 1.0% of the total pumping
pressure, and the pressure drop of the input air would be 1600 Pa
which is tiny compared with the compression ratio.
Influence of Temperature Difference Across Heat Exchanger
[0277] FIGS. 36 and 37 show the efficiencies of a CES as a function
of temperature differences between hot and cold fluids in heat
exchangers for cases with and without superheating, respectively.
Six values of the temperature, 0K, 2K, 4K, 6K, 8K, 10K, are
simulated. The efficiencies of the CES decrease monotonically with
an increase in the temperature difference. When the temperature
difference increases by 1 K, the efficiency of the CES decreases by
.about.0.37% for P1=0.1 MPa without heat recycle, by .about.0.25%
for P1=0.1 MPa with heat recycle, by .about.0.36% for P1=4.0 MPa
without heat recycle, and by .about.1.33% for P1=4.0 MPa with heat
recycle, respectively. Therefore, the temperature difference of hot
and cold fluids in the heat exchangers plays a fairly important
role in the overall performance of the CES.
[0278] The influence of the temperature of the waste heat used for
superheating on the efficiency of the CES for P.sub.1=0.1 MPa and
P.sub.1=4.0 MPa is illustrated in FIG. 38. Five values of the
temperature, 400K, 450K, 500K, 550K, 600K, are simulated. The
selection of the values of the temperature is on the basis of the
available waste heat of different types of power plants. For
example, the temperature of flue gas of a gas turbine power plant
is .about.800 K, the temperature of flue gas of a steam turbine
power plant is .about.400K.about.500K, the temperature of waste
heat from a nuclear power plant is .about.550 K, the temperature of
waste heat from a cement kiln is 700K, and the geothermic
temperature is .about.350K.about.500K.
[0279] The efficiency of the CES increases monotonically with an
increase in temperature of the flue gas containing the waste heat.
If the waste heat temperature increases from 400K to 600K, the
efficiency of the CES increases from 0.558 to 0.749 for P.sub.1=0.1
MPa, and from 0.654 to 1.714 for P.sub.1=4.0 MPa.
[0280] Making the best utilization of waste heat is therefore a
very effective way to improve the performance of the CES. Note that
the waste heat is not accounted for as the input energy, hence the
efficiency can be more than 100%. In addition, the waste heat can
be from geothermal, cement kilns, or other industrial sources.
[0281] FIG. 39 shows the influence of the ambient temperature on
the efficiency of the CES for P.sub.1=0.1 MPa and P.sub.1=4.0 MPa.
Ambient temperatures of 270K, 280K, 290K, 300K and 310K are
simulated. The efficiencies of the ideal cycle for P.sub.1=0.1 MPa,
the practical cycle for P.sub.1=0.1 MPa, the ideal cycle for
P.sub.1=4.0 MPa, and the practical cycle for P.sub.1=4.0 MPa
increase almost linearly with increasing ambient temperature. When
the ambient temperature increases from 270K to 310K, the
efficiencies of the above mentioned cycles increase by 9.7%, 9.1%,
10.2% and 5.5%, respectively. It is therefore concluded that the
CES performs better in locations with a high ambient temperature
such as tropical regions.
Heat Dissipation of Cryogen Tank
[0282] The heat dissipation (leakage) of the cryogenic tank is
about 1% per day in an insulated Dewar at ambient pressure. If more
efforts are taken or cold energy dissipation is utilised, the loss
of efficiency of CES due to the dissipation (leakage) may be less
than 1% per day. This is important when considering the duration of
the energy storage cycle of the CES, i.e. the liquid air must be
used within a certain period of time in order to ensure the overall
efficiency.
[0283] Table 2 shows the process for calculating CES efficiency for
a number of thermodynamic cycles.
TABLE-US-00002 TABLE 2 Cycle 3 Cycle 4 Cycle 5 Cycle 6 Cycle 7
Cycle 8 Cycle 9 Pressure of Liquid air (MPa) 20 20 20 20 20 20 20
Pressure of Input air (MPa) 4.0 4.0 4.0 4.0 0.1 0.1 0.1 Temperature
of working fluid at 400 400 400 400 400 300 300 turbine (K) Ratio
of the input air to the working 0.780 0.780 0.810 0.810 1.69 1.69
1.69 fluid (x) (kg/kg) Ratio of liquid air produced from the 0.730
0.730 0.724 0.724 0 0 0 input air to the working fluid (xy) (kg/kg)
Heat exchanger temperature 0 0 5 5 5 5 5 differences (K) Efficiency
of turbine 100% 88.0% 100% 88.0% 88.0% 88.0% 88.0% Reduction in
efficiency of turbine 0 0 0 0 15.0% 15.0% 15.0% due to
non-isothermicity Efficiency of pump 100% 88.0% 100% 88.0% 88.0%
88.0% 88.0% Efficiency of compressor 100% 88.0% 100% 88.0% 88.0%
88.0% 88.0% Reduction of efficiency of 0 0 0 0 15.0% 15.0% 15.0%
compressor due to non-isothermicity Expansion work (kJ/kg) 615.2
541.4 607.5 534.6 455.6 335.3 335.3 Compression work (Only x kg of
247.2 280.9 256.7 291.7 0 0 0 input air is compressed) (kJ) Pumping
work (kJ/kg) 22.4 25.5 22.4 25.5 25.5 25.5 25.5 Cold energy
recycling (kJ/kg) 0 0 0 0 267.2 267.2 267.2 Net work (Only (1-xy)
kg of working 345.6 235.0 328.4 217.4 697.3 577.0 577.0 fluid is
used in a single cycle) (kJ) Net flowrate of liquid air (based on
0.270 0.270 0.276 0.276 1.0 1.0 1.0 1 kg of working fluid) (kg)
Energy density (E.sub.D) (kJ/kg) 1280.0 870.3 1189.9 787.7 697.3
577.0 577.0 Energy consumed by air liquefaction 1440 1440 1440 1440
1440 1440 1080 (E.sub.C) (kJ/kg) Efficiency of CES
(E.sub.D/E.sub.C) 88.9% 60.9% 82.7% 54.7% 48.4% 40.1% 53.4% Total
amount of exergy (E.sub.I) (kJ/kg) 743 743 743 743 743 743 743
Cycle efficiency (E.sub.D/E.sub.I) (waste heat 172.3% 117.1% 160.1%
106.0% 93.8% 77.7% 77.7% is not included as input energy)
Heat Transfer Analysis--CPS
[0284] The heat exchangers play a critical role in the CPS. The
flow and heat transfer in the heat exchangers in CPS involves three
dimensional, viscous, turbulent and two-phase phenomena. In this
analysis, the following assumptions are made:
(1) Thermodynamic equilibrium between fluid phases (2) Even flow
distribution within heat exchanger (3) Fully developed turbulent
flow (4) Adiabatic shell wall (5) Zero axial conduction (6) No
radiation heat transfer between hot and cold fluids (7) Constant
overall heat transfer coefficient.
Heat Transfer Requirement
[0285] The main CPS system has four heat exchangers and the turbine
uses an additional heat exchanger for isothermal expansion (see
FIG. 4):
(1) Heat exchanger 1 (540): for input air 1 to extract cold from
the working fluid for condensing the input air 1. (2) Heat
exchanger 2 (535): for input air 1 and 4 to extract cold from the
working fluid (3) Heat exchanger 3 (530): input air 1, 3 and 4 to
extract cold from the working fluid (4) Heat exchanger 4 (525): for
input air 1 and 2 to absorb compression heat (5) Heat exchanger 5:
for turbine to absorb heat from atmosphere
[0286] The specific heat transfer requirements for the five heat
exchangers are, respectively:
Heat Exchanger 1 Q.sub.1=x.sub.1(h.sub.6-h.sub.7) Heat Exchanger 2
Q.sub.2=x.sub.1(h.sub.5'-h.sub.6)+x.sub.4 (h.sub.10-h.sub.11) Heat
Exchanger 3
Q.sub.3=x.sub.1(h.sub.5-h.sub.5')+x.sub.3(h.sub.0-h.sub.10)+x.sub.4
(h.sub.0-h.sub.10) Heat Exchanger 4
Q.sub.4=x.sub.1(h.sub.4-h.sub.5)
Heat Exchanger 5 Q.sub.5=T.sub.0(S.sub.0-S.sub.3)
[0287] By using the above equations, the specific heat transfer
requirements for Q1 to Q5 are 47.9 kJ/kg, 165.9 kJ/kg, 225.4 kJ/kg,
57.8 kJ/kg and 597.7 kJ/kg respectively, where the ambient
temperature is 300 K. The maximum specific heat transfer
requirement of the entire CPS system, therefore, is 1102.0 kJ/kg.
In the following sections, analyses will be based on this value of
heat transfer requirements.
Preliminary Design of Heat Exchanger
[0288] The following is an estimation of the size of the heat
exchangers based on the performance of the plate and fin type. The
overall average heat transfer coefficient is taken as 100
W/m.sup.2K; the average temperature difference between hot and cold
fluids, .DELTA. T, is assumed as 2 K; the ratio of compactness,
.theta., is taken as 1000 m.sup.2/m.sup.3. The compactness could be
much higher, so the estimation is on the conservative side. The
maximum heat transfer requirements, H.sub.R, is given as 1102.0
kJ/kg on the basis of the above calculation. For a CPS with a work
output of 1 kW, the heat transfer requirement is given by:
Q = 1 H R W R E E = 2.184 kW ##EQU00098##
where W.sub.R and E.sub.E are the maximum specific work of liquid
air and the energy efficiency of CPS, respectively. The size of the
heat exchanger for a 1 kW work output can be calculated by:
V = Q U _ .DELTA. T _ = 0.011 m 3 . ##EQU00099##
If a factor of safety is given as 4, the size of the heat exchanger
for a unit work output would be 0.044 m.sup.3.
[0289] The liquid nitrogen viscous pressure drop is reported to be
about 0.5 bar and the pressure drop of input air was about 400 Pa.
If a safety factor of 4 is used, then the liquid air pressure drop
would be about 2 bar which is about 1.0% of the total pumping
pressure (200 bar), and the pressure drop of the input air would be
1600 Pa (0.016 bar) which is very small compared with the
compression ratio (.about.0.2% of 8 bar).
Influence of Temperature Difference Across Heat Exchanger
[0290] FIG. 40 shows the efficiency of a CPS as a function of
temperature difference between hot and cold fluids in heat
exchangers. Six values of the temperature, 0K, 2K, 4K, 6K, 8K, 10K,
are simulated. The efficiency of the CPS decreases monotonically
with an increase in the temperature difference. When the
temperature difference increases by 1 K, the efficiency of the CPS
decreases by 0.4%. Therefore, the temperature difference of hot and
cold fluids in the heat exchangers plays a fairly important role in
the overall performance of the CPS.
Heat Dissipation of Cryogen Tank
[0291] The heat dissipation (leakage) rate of the cryogenic tank is
about 1% per day in an insulated Dewar at the ambient pressure. If
more efforts are taken or cold energy dissipation is utilised, for
example for air conditioning, the loss of efficiency of CPS due to
the dissipation (leakage) may be less than 1% per day. The
efficiency of heat dissipation as a function of time for four
dissipation rates of 1%, 0.75%, 0.50%, 0.25% per day, is shown in
FIG. 41. The efficiency of heat dissipation E.sub.dis is defined
as
E dis = M ideal M ac ##EQU00100##
where M.sub.ideal refers to the total amount of mass of liquid air
without dissipation, and M.sub.ac is the actual total amount of
mass of liquid air with the dissipation. It can be seen that the
efficiency of heat dissipation decreases with increasing time and
dissipation rate. This indicates that the CPS operation should be
within a certain period of time in order to ensure a good overall
efficiency. It is essential to decrease the heat dissipation
especially for a long term journey. For a dissipation rate of
.about.0.5% per day, the total loss over a duration of 30 days is
.about.7.5%.
Example of a Lab Scale CES System
[0292] An exemplary small lab scale CES system with a capacity of
100 kWh is illustrated schematically in FIG. 42. This represents a
system at a scale much smaller than the probable size of commercial
units and is designed for testing operating parameters and
optimising performance of the system. A fill scale CES system may
contain additional components which are not included in the lab
scale system. The system has a 12.5 kW power rating and an 8 hour
discharge time. The power rating could also be suitable for the
power needs of multiple households in a microgeneration
configuration. The 8 hour discharge time (100 KWh stored) is chosen
because this is near to the maximum discharge duration required for
energy storage applications suggested by bodies such as the Sandia
laboratories.
[0293] The experimental system consists of 8 major components, a
cryogenic tank 600, a pump 610, a heat exchanger 620, a turbine
630, a transmission box 640, a blower 650, a drier 660 and a
three-way valve 670. The system works as follows:
1) Liquid air (working fluid) from a cryogen plant or a storage
depot is fed into the cryogenic tank 600. 2) Working fluid is
pumped and heated before flowing into the turbine 630, where it
expands to produce power to drive the blower 650. The blower 650
has two functions, one is to provide the input air for recovery of
the cold energy through the heat exchanger 620, and the other one
is to provide a load to the turbine 630 (acting as a generator). 3)
A small fraction of the air from the blower 650 (input air) is
introduced to the heat exchanger 620 via the three-way valve 670
and the drier 660. 4) No liquid air is produced in the lab scale
system to reduce the capital cost. This, however, does not affect
assessment of the CES performance as the measured data are
sufficient for such a purpose.
Thermodynamic Analyses
[0294] The thermodynamic cycle of the lab scale CES system is shown
in FIG. 43. Let T.sub.0 h.sub.0 and S.sub.0 denote respectively the
ambient temperature, enthalpy and entropy, the processes and their
heat, work and exergy are given in the following:
[0295] 1) 1-2: Pumping of working fluid: The working fluid (liquid
air) from the cryogenic tank is pumped from the ambient pressure
P.sub.0 to P.sub.2. The specific work done on the liquid air
is:
W 1 - 2 = V 1 ( P 2 - P 0 ) = ( P 2 - P 0 ) .rho. 1 .
##EQU00101##
The above work can also be expressed by the enthalpy difference
between state 2 and state 1: W.sub.1-2=h.sub.2-h.sub.1. The total
cold exergy (maximum work availability) of the working fluid at
state 1 is:
Ex.sub.1=T.sub.0(S.sub.0-S.sub.1)-(h.sub.0-h.sub.1).
[0296] 2) 2-3: Isobaric heating of working fluid: The working fluid
is heated by the input air from T.sub.2 to T.sub.3. The specific
work done in this process is zero: W.sub.2-3=0. The specific heat
absorbed from the input air is: Q.sub.2-3=h.sub.3-h.sub.2. The
exergy released in the process 2-3 is:
Ex.sub.2-3=T.sub.0(S.sub.3-S.sub.2)-(h.sub.3-h.sub.2).
[0297] 3) 3-4: Expansion of working fluid: The working fluid with a
high pressure expands in the turbine to deliver work. If an ideal
isothermal process is considered, the specific work done in the
process is: W.sub.3-0=T.sub.0(S.sub.0-S.sub.3)-(h.sub.0-h.sub.3).
The specific heat absorbed from the ambient in an ideal isothermal
process is: Q.sub.3-0=T.sub.0(S.sub.0-S.sub.3). If the expansion of
the working fluid is adiabatic, the specific ideal work W.sub.ad
will be:
W ad = k k - 1 RT 3 [ P 0 P 2 ) ( k - 1 ) k - 1 ] ##EQU00102##
and there is no heat absorption in the process: Q.sub.ad=0. The
actual work, however, is expected to be in the range between
W.sub.3-0 and W.sub.ad. A factor called isothermicity, .gamma., is
often used as an index, which is defined as the ratio of the actual
work to the isothermal work:
.gamma. = W 3 - 4 W 3 - 0 . ##EQU00103##
Thus, the actual work W.sub.3-4 can be expressed as:
W.sub.3-4=.gamma.W.sub.3-0=.gamma.[T.sub.0(S.sub.0-S.sub.3)-(h.sub.0-h.su-
b.3)].
[0298] 4) 6-7: Extraction of cold energy of the working fluid by
input air isobarically: The cold energy of the working fluid is
extracted by the input air isobarically through the heat exchanger.
The specific work done in this process is zero: W.sub.6-7=0. The
cold energy from the working fluid in the process 6-7 is:
Q.sub.6-7=h.sub.6-h.sub.7. The exergy obtained by the input air
over the process is given by:
Ex.sub.6-7=T.sub.6(S.sub.6-S.sub.7)-(h.sub.6-h.sub.7). From the
above analysis, the specific ideal net work output of the cycle
should be:
W net = W 3 - 4 - W 1 - 2 + F 2 F 1 Ex 6 - 7 = .gamma. T 0 ( S 0 -
S 3 ) - ( h 0 - h 3 ) - ( h 1 - h 2 ) + F 2 F 1 [ T 6 ( S 6 - S 7 )
- ( h 6 - h 7 ) ] ##EQU00104##
where F.sub.1 and F.sub.2 are respectively the flowrates of the
working fluid and input air, respectively.
[0299] The efficiency of the lab scale CES experimental system can
therefore be expressed by:
E C = W net Ex 1 = W 3 - 4 - W 1 - 2 + F 2 F 1 Ex 6 - 7 Ex 1 =
.gamma. T 0 ( S 0 - S 3 ) - ( h 0 - h 3 ) - ( h 1 - h 2 ) + F 2 F 1
[ T 6 ( S 6 - S 7 ) - ( h 6 - h 7 ) ] T 0 ( S 0 - S 1 ) - ( h 0 - h
1 ) ##EQU00105##
where Ex.sub.1 is the total cold exergy contained in the working
fluid. In the actual experimental system, a certain amount of work
is needed to pump the input air through the heat exchanger;
therefore the work of process 6-7 is not zero. The actual specific
net work output should be:
W net = W 3 - 4 - W 1 - 2 + F 2 F 1 Ex 6 - 7 - F 2 F 1 W 6 - 7 =
.gamma. T 0 ( S 0 - S 3 ) - ( h 0 - h 3 ) - ( h 1 - h 2 ) + F 2 F 1
[ T 6 ( S 6 - S 7 ) - ( h 6 - h 7 ) ] - F 2 F 1 W 6 - 7 .
##EQU00106##
The efficiency of the lab scale CES system therefore becomes
E C = W net Ex 1 = W 3 - 4 - W 1 - 2 + F 2 F 1 Ex 6 - 7 - F 2 F 1 W
6 - 7 Ex 1 = .gamma. T 0 ( S 0 - S 3 ) - ( h 0 - h 3 ) - ( h 1 - h
2 ) + F 2 F 1 [ T 6 ( S 6 - S 7 ) - ( h 6 - h 7 ) ] - F 2 F 1 W 6 -
7 T 0 ( S 0 - S 1 ) - ( h 0 - h 1 ) ##EQU00107##
Measurement Techniques and Data Processing
[0300] A suitable measurement system is shown schematically in FIG.
42. There are 20 measurement channels with 7 for thermocouples, 7
for pressure transducers, 2 for flow rates, 1 for electric voltage,
1 for electric current, and I for torque/speed. A data acquisition
system is linked to a computer for data acquisition, storage and
processing. The measurement channels comprise: [0301] (1) T1:
Temperature of the working fluid at the inlet of the pump 610.
[0302] (2) T2: Temperature of the working fluid at the outlet of
the pump 610/the inlet of the heat exchanger 620. [0303] (3) T3:
Temperature of the working fluid at the outlet of the heat
exchanger 620/inlet of the turbine 630. [0304] (4) T4: Temperature
of the working fluid at the outlet of the turbine 630. [0305] (5)
T5: Temperature of the air at the inlet of the blower 650 (ambient
temperature). [0306] (6) T6: Temperature of the input air at the
inlet of the heat exchanger 620. [0307] (7) T7: Temperature of the
input air at the outlet of the heat exchanger 620. [0308] (8) P1:
Static pressure of the working fluid at the inlet of the pump 610.
[0309] (9) P2: Static pressure of the working fluid at the outlet
of the pump 610/inlet of the heat exchanger 620. [0310] (10) P3:
Total pressure of the working fluid at the outlet of the heat
exchanger 620/inlet of the turbine 630. [0311] (11) P4: Total
pressure of the working fluid at the outlet of the turbine 630.
[0312] (12) P5: Total pressure of the air at the inlet of the
blower 650 (ambient pressure). [0313] (13) P6: Static pressure of
the input air at the inlet of heat exchanger 620. [0314] (14) P7:
Static pressure of the input air at the outlet of heat exchanger
620. [0315] (15) F1: Flow rate of the working fluid delivered by
the pump 610. [0316] (16) F2: Flow rate of the input air through
the heat exchanger 620. [0317] (17) V1: Electric voltage of the
pump 610. [0318] (18) C1: Electric current of the pump 610. [0319]
(19) .omega.1: Rotary speed of the turbine 630. [0320] (20) M1:
Output torque of the turbine 630.
[0321] From the above thermodynamic analyses, it can be seen that
seven variables are needed to obtain the actual efficiency of the
lab scale experimental system, including W.sub.3-4, W.sub.1-2,
Ex.sub.6-7, W.sub.Blower, Ex.sub.1, F.sub.1 and F.sub.2. The
methodologies for obtaining these parameters follow:
(1) W.sub.3-4: actual work output of the turbine: A torque/speed
meter is directly connected to the axis of the turbine 630, and the
blower 650 is used as a load. The work output of the turbine 630 is
obtained by multiplying the measured torque (M.sub.1) and speed
(.omega..sub.1): W.sub.3-4=M.sub.1.omega..sub.1. (2) W.sub.1-2:
work consumed by the pump: In the experimental system, the pump 610
is driven by a motor. The actual work consumed by the pump 610 can
therefore be obtained by measuring the electric voltage (V.sub.1)
and current (C.sub.1) of the motor: W.sub.1-2=V.sub.1C.sub.1. The
result of W.sub.1-2 accounts for both the efficiencies of the pump
and the motor 610. (3) Ex.sub.6-7: cold exergy recycled by input
air: The cold exergy recovered by the input air can be calculated
by: Ex.sub.6-7=T.sub.6(S.sub.6-S.sub.7)-(h.sub.6-h.sub.7). To
obtain the entropies and enthalpies of the input air, i.e. S.sub.6,
S.sub.7, h.sub.6 and h.sub.7, two thermocouples and two pressure
transducers are used in the experimental system at the inlet and
outlet of the heat exchanger 620, respectively. Using the measured
data of T.sub.6, T.sub.7, P.sub.6 and P.sub.7, the entropies and
enthalpies of the input air can be found from the thermodynamics
data tables for the air. (4) W.sub.6-7: work needed for pumping the
input air: The specific work consumed for pumping the input air is
calculated by the pressure difference between the inlet and outlet
of the heat exchanger 620: W.sub.6-7=P.sub.7-P.sub.6.
[0322] (5) F.sub.1: Flow rate of working fluid: The flow rate of
the working fluid is measured by the flow meter installed at the
inlet of the pump 610.
(6) F.sub.2: Flow rate of input air: The flow rate of the input air
is measured by the flow meter installed at the outlet of the heat
exchanger 620. (7) Ex.sub.1: total cold exergy contained in working
fluid: The total cold exergy recovered from the working fluid is
calculated by: Ex.sub.1=T.sub.0(S.sub.0-S.sub.1)-(h.sub.0-h.sub.1).
To obtain the entropies and enthalpies of the working fluid, i.e.
S.sub.0, S.sub.1, h.sub.0 and h.sub.1, two thermocouples and two
pressure transducers are installed at the inlet and outlet of the
heat exchanger 620, respectively. Using the data of T.sub.5,
T.sub.1, P.sub.5 and P.sub.1, the entropies and enthalpies of the
working fluid can be obtained by referring to the thermodynamic
data tables for the air.
[0323] Parameters related to individual components that can be
obtained from the experimental CES include:
(1) Cryogenic Tank
[0324] a. The volume of liquid air can be obtained from a level
indicator; the heat dissipation can be calculated from the
difference in volume over a known time period. [0325] b. The
temperature at the outlet of the cryogenic tank 600 (T.sub.1)
[0326] c. The pressure at the outlet of the cryogenic tank 600
(P.sub.1) [0327] d. The flow rate of the working fluid
(F.sub.1)
(2) Pump
[0327] [0328] a. The flow rate of the pump 610 (F.sub.1) [0329] b.
The temperature at the inlet (T.sub.1) and outlet (T.sub.2) of the
pump 610 [0330] c. The pressure at the inlet (P.sub.1) and outlet
(P.sub.2) of the pump 610 [0331] d. The efficiency of the pump
610:
[0331] .eta. p = F 1 .DELTA. P t V 1 C 1 = F 1 ( P 2 - P 1 ) V 1 C
1 ##EQU00108##
(3) Heat Exchanger
[0332] a. The temperature of the working fluid at the inlet
(T.sub.2) and outlet (T.sub.3) of the heat exchanger 620 [0333] b.
The pressure of the working fluid at the inlet (P.sub.2) and outlet
(P.sub.3) of the heat exchanger 620 [0334] c. The temperature of
the input air at the inlet (T.sub.6) and the outlet (T.sub.7) of
the heat exchanger 620 [0335] d. The pressure of the input air at
the inlet (P.sub.6) and outlet (P.sub.7) of the heat exchanger 620
[0336] e. The flow rate of the working fluid (F.sub.1) [0337] f.
The flow rate of the input air (F.sub.2) [0338] g. The temperature
differences of the working fluid and the input air between the
inlet and outlet of the heat exchanger 620: (T.sub.7-T.sub.2) and
(T.sub.6-T.sub.3) [0339] h: The pressure differences of the working
fluid and the input air between the inlet and outlet of the heat
exchanger 620: (P.sub.7-P.sub.2) and (P.sub.6-P.sub.3)
(4) Turbine
[0339] [0340] a. The temperature of the working fluid at the inlet
(T.sub.3) and the outlet (T.sub.4) of the turbine 630 [0341] b. The
pressure of the working fluid at the inlet (P.sub.3) and the outlet
(P.sub.4) of the turbine 630 [0342] c. The output torque (M.sub.1)
and rotary speed (.omega..sub.1) of the turbine 630 [0343] d. The
efficiency of the turbine 630 calculated by:
[0343] .eta. T = M 1 .omega. 1 F 1 ( P 3 - P 4 ) ##EQU00109##
[0344] e. The isothermicity of the expansion in the turbine 630
calculated by:
[0344] .gamma. = M 1 .omega. 1 T 0 ( S 0 - S 3 ) - ( h 0 - h 3 )
##EQU00110##
(5) Blower
[0345] a. The temperature of air at the inlet (T.sub.5) and outlet
(T.sub.6) of the blower 650 [0346] b. The pressure of air at the
inlet (P.sub.5) and outlet (P.sub.6) of the blower 650 [0347] c.
The input torque (M.sub.1) and rotary speed (.omega..sub.1) of the
blower 650 (driven by the turbine 630).
Detailed Thermodynamic Analyses of the Components of the Small
Scale Lab CES
[0348] (1) Cryogenic tank: The flow rate of the fuel (liquid air)
can be calculated by:
F l = P o .eta. E D .rho. l ##EQU00111##
where F.sub.l, P.sub.o, .eta., E.sub.D and .rho..sub.l are the
flowrate of liquid air, power of the system, efficiency of the
turbine 630, energy density of liquid air and density of liquid
air, respectively. The volume of the fuel tank 600 is given by:
V l = S f F l O t E dis ##EQU00112##
where S.sub.f, V.sub.l, O.sub.t, E.sub.dis are respectively the
safe factor, volume of liquid air, operating duration and
efficiency of heat dissipation of the tank. If a cubic tank is
assumed, the length of each side, d, is V.sub.1.
[0349] If the working pressure of the working fluid is 20 MPa, the
ambient temperature is 300 K, the ideal specific energy density of
liquid air is .about.455 kJ/kg, the density of liquid air at the
ambient pressure is .about.876 kg/m.sup.3, the efficiency of the
turbine 630 is 0.8 and the total power of the lab scale
experimental system is 12.5 kW, the flow rate of liquid air is:
F l = P o .eta. E D .rho. l = 12.5 0.8 455 876 = 141.0 l / h .
##EQU00113##
If a safe factor of 1.3 is considered and the efficiency of heat
dissipation is taken as 0.95, then the volume of the cryogenic tank
600 for a total capacity of 100 kWh is:
V l = S f F l O t E dis = 1.3 .times. 0.141 .times. 8 0.95 = 1.55 m
3 . ##EQU00114##
If a cubic tank is assumed, the length of each side is 1.14 m.
[0350] Due to heat transfer, liquid air evaporates in the cryogenic
tank 600 and the pressure of liquid air at the outlet of the tank
600 (inlet of the pump 610) is higher than the ambient pressure
which leads to a decrease in the work consumed by the pump 610.
Given that self-pressurisation of the cryogenic tank 600 is
unavoidable, a safety valve is included to relieve the pressure
once it exceeds a certain level. It is possible to control the tank
pressure through alternative systems a safety valve.
[0351] (2) Pump: Key parameters associated with the pump include
the working fluid flow rate, inlet pressure, outlet pressure,
working temperature and power consumption. The flow rate of the
pump is the same as that for the cryogenic tank: F.sub.l=141.0 l/h.
The inlet pressure of liquid air is determined by the outlet
pressure of the cryogenic tank. As a safety valve is used in the
lab scale system, the pressure of the tank cannot be determined a
priori. However, the cryogen pump can work over a certain range of
inlet pressures at a given outlet pressure. Therefore, the inlet
pressure of the pump is taken as P.sub.1=0.1.about.3.0 MPa. The
outlet pressure of the cryogen pump is equal to the working
pressure of the working fluid which is given as 20 MPa. Therefore,
P.sub.2=20 MPa . The cryogen pump should work in the normal
laboratory temperature. Therefore the working temperature is
selected as 0.degree. C..about.40.degree. C. The temperature of the
working fluid at the inlet of the pump is approximately the boiling
point of liquid air (-196.degree. C.). The temperature of the
working fluid at the outlet of pump is expected to be
.about.192.degree. C. after an adiabatic pressurisation process.
The power consumed by the pump is determined by its efficiency
given the outlet pressure and flow rate. If the efficiency of the
pump is assumed as 0.8, the power requirement of the pump is:
P pump = F l .rho. l W 1 - 2 .eta. p = 1.0 kW . ##EQU00115##
If a safe factor of 1.5 is used for the motor of the cryogen pump,
the power of the motor will be 1.5 kW.
[0352] (3) Heat Exchanger: Key parameters associated with the heat
exchanger include working pressures, flow rates and pressure drops
of both the working fluid and the input air, and temperatures of
the working fluid and input air at the inlet and outlet of the heat
exchanger. The working pressure of the working fluid is
approximately equal to the outlet pressure of the pump: P.sub.2=20
MPa. The working pressure of the input air should be close to the
ambient pressure to minimise the work consumed by the blower:
P.sub.7=P.sub.0. The inlet pressure of the input air is
approximately equal to the pressure drop across the heat exchanger:
P.sub.6=P.sub.Loss+P.sub.7. The flow rate of the working fluid has
been given above: F.sub.1=141.0 l/h=(123 kg/h). The flow rate of
the input air is influenced by the performance of the heat
exchanger. An approximate value is obtained by thermodynamic
calculation as F.sub.2=206.0 kg/h. The pressure drop of the working
fluid across the heat exchanger depends on the engineering design
of the heat exchanger. It is estimated, however, to be of an order
of .about.1000 Pa. The pressure drop of the input air across the
heat exchanger also depends on the design. It is also estimated to
be .about.1000 Pa. The temperature of the working fluid at the
inlet of the heat exchanger is approximately equal to that at the
outlet of the pump if the heat loss of pipes/joints/valves etc is
neglected, i.e. T.sub.2=-192.degree. C. The temperature of the
working fluid at the outlet of the heat exchanger depends on the
performance of the heat exchanger, it is estimated to be close to
the ambient temperature with a temperature difference assumed (i.e.
5.degree. C.), i.e. T.sub.3=22.degree. C. The temperature of the
input air at the inlet of the heat exchanger is approximately the
ambient temperature. The temperature of the input air at the outlet
of the heat exchanger also depends on the performance of the heat
exchanger; but is estimated to be close to the temperature of the
working fluid at the inlet of the heat exchanger
(.about.-192.degree. C.).
[0353] (4) Turbine: In analysing the performance of the turbine a
multistage adiabatic expansion process with inter-heating is
considered. The pressure of the working fluid at the inlet of the
turbine has been given above as P.sub.3=20 MPa. The temperature of
the working fluid at the inlet of the turbine should be close to
the ambient temperature after being heated by the heat exchanger.
If a temperature difference of 5.degree. C. (below ambient) is
considered, the temperature of the working fluid at the inlet of
the turbine is 22.degree. C. (ambient temperature taken as 300K):
T.sub.3=22.degree. C.
[0354] The number of stages is a key parameter of the turbine; more
stages mean nearer isothermal operation hence more work output (see
FIG. 44). However, more stages also mean more mechanical
complexity, high pressure loss, and a high cost. A balance between
the two is needed. Construction of FIG. 44 is based on the
following assumptions:
[0355] Ideal Case: The pressure of the working fluid at the inlet
of the turbine is 20 MPa, the efficiency of the turbine is 100%,
and the temperature of the working fluid at the inlet of each stage
is 27.degree. C.
[0356] Practical Case: The pressure of the working fluid at the
inlet of the turbine is 20 MPa, the efficiency of the turbine is
89%, and the temperature of the working fluid at the inlet of each
stage is 22.degree. C.
[0357] Both the ideal and practical work outputs of the turbine
increase with increasing number of stages and level off between 4
to 8 stages. The total number of stages is also limited by the
maximum expansion ratio of the turbine, which is normally less than
3.0. FIG. 45 shows the expansion ratio of each stage as a function
of the number of stages of the turbine. It can be seen that the
expansion ratio exceeds 3 if the number of stages is less than 4.
As a consequence, the number of the stages of the turbine should be
more than 4. Consequently, the number of stages should lie between
4 and 8.
[0358] The pressure of the working fluid at the outlet is generally
a little higher than the ambient pressure to ensure the working
fluid flows smoothly. The pressure of the working fluid at the
outlet is often selected as .about.0.13 MPa. If the number of
stages is 6 and the temperature of the working fluid at the inlet
is 22.degree. C., the temperature of the working fluid at the
outlet is approximately -44.degree. C. Air at such a temperature
can be recycled for liquid air production in large CES systems. It
could also be used for industrial freezing and air conditioning in
summer. The flow rate of the working fluid is equal to the flow
rate of the pump: 123 kg/h (141 l/hr). Due to the low flow rate and
high pressure of the working fluid, the size of the first stage of
the turbine will be several millimetres, which is classified as a
micro turbine.
[0359] (5) Blower: Key parameters associated with the blower are
the pressure, flow rate, power and efficiency. The rated power
should be approximately equal to the work output of the turbine
(.about.12.5 kW), and the pressure should be higher than the
pressure drop of the input air across the heat exchanger.
Selection of Suitable Components
[0360] The following component selection is based on the analyses
detailed above.
[0361] (1) Cryogenic tank: Product No. C404C1 (Model ZCF-2000/16)
of Si-Chuan Air Separation Plant (Group) Co. Ltd is a suitable
vertical type cryogenic tank having a double-walled and vacuum
powder insulated structure; see FIG. 46 for a schematic diagram.
This cryogenic tank has the following parameters: [0362]
Capacity=2000 litres [0363] Maximum Working pressure=1.6 MPa [0364]
Empty Tank Weight=2282 kg [0365] Dimensions
(Dia.times.H).times.1712 mm.times.3450 mm [0366] Daily boil-off
(percentage of liquid air evaporated per day at 20.degree. C. and
0.1 MPa)=<0.96%.
[0367] (2) Pump: A reciprocating piston cryogenic liquid pump is
recommended for the lab scale CES experimental system and Product
No. B228 of the Cryogenic Machinery Corporation (a Si-Chuan Air
Separation Plant (Group) Co. Ltd company) is suitable. This pump
has a high vacuum insulated pump head, which can reduce
vaporisation loss and the suction pressure of pump. The piston ring
and filling ring of the pump use non-metallic cryogenic material
possessing good plasticity and lubricating ability. The use of
special lubricant ensures that the pump can work for combustible or
even explosive liquids such as liquid oxygen. The internal
structure of the pump is shown in FIG. 47. This cryogenic pump has
the following parameters: [0368] Working Fluid=Liquid
air/oxygen/nitrogen/argon [0369] Inlet Pressure=0.05.about.1.5 MPa
[0370] Outlet Pressure=20.about.35 MPa [0371] Flow
rate=50.about.150 l/h [0372] Power=3.0 kW [0373] Working
Temperature=-10.about.40.degree. C. [0374] Weight=150 kg.
[0375] (3) Heat Exchanger: The heat exchanger works at a high
pressure of 20 MPa and across a very large temperature difference
(-196.degree. C..about.27.degree. C.). The flow rate of the working
fluid is 123 kg/h. No existing products have been found that are
suitable for the purpose. Therefore, a specially designed and
fabricated heat exchanger is needed. Such a heat exchanger could be
a tube-fin structure enclosed in a shell with the following
parameters: [0376] Working Fluid=Liquid air [0377] Heating
Fluid=Ambient Air [0378] Pressure of (cold) working fluid=20 MPa
[0379] Pressure of Heating Fluid=0.1 MPa [0380] Flow rate of
working fluid=123 kg/h [0381] Flow rate of heating fluid=.about.206
kg/h [0382] Pressure loss of working fluid=<500 Pa [0383]
Pressure loss of heating fluid=<1000 Pa [0384] Working
Temperature=-10.about.40.degree. C. [0385] Material of tube=304
Stainless Steel [0386] Material of fin and
shell=Stainless/Aluminium Alloy [0387] Dimensions
(Length/Width/Height)=2.5 m/2.2 m/0.8 m [0388] Weight=.about.1200
kg.
[0389] (4) Turbine: The performance of the turbine plays a dominant
role in the performance of the whole lab scale system. The output
work of a turbine is normally used to drive a motor, a compressor,
a fan, or a power generator. As the inlet pressure of the proposed
turbine is high (.about.20 MPa) and the flow rate of working fluid
is low (.about.123 kg/h), the turbine has to be a micro-turbine
with a diameter of several millimetres. FIG. 48 shows a schematic
diagram of a suitable turbine. However, no existing turbines have
been found that are compatible with the proposed lab scale system.
Therefore, a specially designed and fabricated turbine is
needed.
[0390] (5) Blower: The blower should be able to deliver a total
pressure to overcome the pressure drop of the input air. As the
blower also acts as a load of the turbine, it must be rated at a
total power approximately equal to the work output of the turbine
(-12.5 kW). A blower such as Beijing Dangdai Fan Company's mixed
flow GXF-C (product code No. 6.5-C) is suitable. This blower has
the following parameters: [0391] Working Fluid=Air [0392] Total
Pressure=1162 Pa [0393] Flow rate=24105 m.sup.3/h [0394] Rotary
Speed=2900 rpm [0395] Noise=83 dB(A) [0396] Power=15 kW [0397]
Dimensions (Length/Width/Height)=0.845 m/0.751 m/0.800 m [0398]
Weight=234 kg
[0399] (6) Other Components: As the rotary speed of the turbine is
normally very high (tens of thousands of rpm), whereas the
rotational speed of the proposed blower is low (2900 rpm), a
transmission system is necessary for the small scale CES
experimental system. In addition, to avoid icing of water (from the
input air) on the wall of the heat exchanger, a drier is necessary
to dehumidify the input air before it enters the heat
exchanger.
[0400] Component Integration Liquid air from a cryogen plant is
transported to the laboratory by a cryogenic truck and fed into the
cryogenic tank C404C1. The reciprocating piston cryogenic liquid
pump B228 pressurises the liquid air and provides kinetic energy
for the working fluid to flow through the heat exchanger. The
working fluid is heated in the heat exchanger by the input air
provided by the blower GXF-C-6.5C, which also serves as the load of
the micro-turbine in which the working fluid expands to provide
power of the blower. Only a fraction of air from the blower is used
as the input air.
Technological and Economical Comparisons of CES with Other Storage
Systems
[0401] Currently existing energy storage systems will now be
evaluated and compared with the CES. The data of the CES is
calculated based on a 500 MWh storage volume and a discharge time
of 8 hours. The data for other energy storage systems are mainly
taken from J. Kondoh et al. "Electrical energy storage systems for
energy networks" (2000, Energy Conversion & Management, vol.
41, 1863-1874), P. Denholm et al. "Life cycle energy requirements
and greenhouse gas emissions from large scale energy storage
systems" (2004, Energy Conversion and Management, vol. 45,
2153-2172), and F. R. Mclarnon et al. "Energy storage" (1989,
Annual Review of Energy, vol. 14, 241-271).
[0402] Output power and output duration: The relationship between
the output power and the output duration of the storage systems is
shown in FIG. 49. Each storage system has a suitable range, and
they can be classified into two types: the daily load levelling
type and the electric power quality improving type.
[0403] Pumped hydro, CAES, batteries and CES are suitable to level
daily load fluctuation. The superconducting magnet and flywheel
with conventional bearing have a fast response and, therefore, can
be utilised for the instantaneous voltage drop, flicker mitigation
and short duration UPS.
[0404] Other systems such as the flywheel with levitation bearing,
double-layer capacitor and redox supercapacitor are promising for
small capacity energy storage and short output duration (less than
1 h).
[0405] The output power and duration of the CES is better than
batteries, competitive with CAES and slightly lower than the pumped
hydro. However, as discussed before, the pumped hydro requires
special geographical location. Furthermore, as discussed below,
pumped hydro requires a very high capital cost.
[0406] The relationship between the efficiency and the cyclic
period is shown in FIG. 50. The downwards concave curves are due to
self-discharge or energy dissipation. The efficiency of the CES
without superheat is lower than other energy storage systems.
However, if the waste heat is recycled to superheat the working
fluid in CES, its efficiency is competitive with other energy
storage systems. Furthermore, the efficiency of CES with
superheating increases with improvement of the air liquefaction
process as discussed above.
[0407] The energy storage densities of different energy storage
systems are shown in FIG. 51. The data is based on the following:
[0408] The energy stored in a pumped hydro plant is calculated
based on mgh, where m is the mass of water, g is acceleration due
to gravity, and h is the effective head which is assumed to be 500
m. [0409] The cavern volume of CAES is assumed to be 54,000 m.sup.3
at about 60 atm. The stored air allows the plant to produce 100 MW
for 26 h continuously. The calculation of the energy density of
CAES does not include the volume of fuel storage, motor/generator,
compressor and expanders. [0410] Calculation of the energy density
of the CES is based on the stored energy and the volume of the
cryogen tank and heat exchangers; the volume of the
motor/generator, compressor and expanders are not considered as
they are at least an order of magnitude smaller than the cryogen
tank. [0411] The energy density for other systems is calculated by
dividing the output power by the volume of the storage device.
[0412] It can be seen that the CES and advanced secondary Na/S
batteries have the highest energy densities among all systems. The
energy density of the CES is higher than CAES by more than an order
of magnitude and higher than pumped hydro by about two orders of
magnitude.
[0413] The lifetimes of storage systems are shown in Table 3. The
cycle durability of secondary batteries is not as high as other
systems owing to the chemical deterioration with the operating
time. Many of the components in CES are similar to those used in
CAES. Therefore it is expected that the CES will have a similar
life time to CAES.
TABLE-US-00003 TABLE 3 Life time of the electrical energy storage
systems Systems Years Cycles Pumped Hydro 40~60 Almost unlimited
CAES 20~40 Almost unlimited CES 20~40 Almost unlimited Lead Acid
Battery 10~15 2000 Na/S battery 10~15 2000~2500 Zn/Br Battery 10
1500 Redox Flow Battery 10,000 Flywheel >15 >20,000
Double-layer capacitor >50,000 Redox super capacitor 5
[0414] FIG. 52 shows the relationship between the output power per
unit capital cost and the storage energy capacity per unit capital
cost of the compared systems. It can be seen that the CAES has the
lowest capital cost per unit output power of all the systems. The
capital cost of the advanced batteries (Na/S, Zn/Br, and vanadium
redox flow) is slightly higher than the breakeven cost against the
pumped hydro although the gap is gradually closed. The SMES and
flywheel are suitable for high power and short duration
applications since they are cheap on the output power basis but
expensive in terms of the storage energy capacity.
[0415] In terms of the CES, the output adjusted capital cost of the
CES is lower than that of the CAES because the life time of the CES
is equal to that of the CAES, the initial investment of the CES is
less than that of the CAES as no cavern is needed, and the energy
density of the CES is higher than that of the CAES by at least an
order of magnitude.
[0416] Therefore, the capital cost of the CES is lowest of all of
the systems examined. In addition, the CES offers a flexibility in
terms of commercial operations as products such as oxygen, nitrogen
and argon can also be produced.
[0417] Construction of a pumped hydro storage system inevitably
involves the destruction of trees and green land for in order to
build the reservoirs. The construction of the reservoirs could also
change the local ecological system which also presents
environmental consequences. CAES is based on conventional gas
turbine technology and involves the combustion of fossil fuel and
consequently the emission of contaminates, whilst secondary
batteries produce solid toxic waste.
[0418] However, CES is benign to the environment. For example,
CO.sub.2 and SO.sub.x are removed during the liquefaction process,
which help with mitigating the negative environmental issues
associated with the burning of fossil fuels. Undesirable airborne
particulates are also removed during production of liquid air.
[0419] Therefore it can be concluded that CES has a better
performance than other energy storage systems in terms of energy
density, lifetime, capital cost and environmental impact. It is
very competitive in comparison to other systems in terms of the
output power and duration and energy efficiency. Compared with
cryogenic engines for vehicles, the work output and efficiency of
CES are much higher due to the use of both `heat` and `cold`
recycles. The optimal pressure of the working fluid is .about.20
MPa for the CES. The optimal pressure of the input air is found to
be .about.0.1 MPa when there is no waste heat recycled. However,
when waste heat is used, the optimal input pressure could be either
0.1 MPa or 4.0 MPa. Based on an efficiency of 0.4 kWh/kg for air
liquefaction, an overall efficiency of the CES operated in an ideal
cycle is estimated at 0.516 for cases without using the waste heat
recycle, and at 0.612 for cases using waste heat from flue gas at a
temperature of 127.degree. C. If the efficiency of the air
liquefaction is taken as 0.3 kWh/kg, then the overall efficiency of
the CES operated in an ideal cycle would be 0.688 for cases which
do not use the waste heat recycle and at 0.816 for cases which do
use the waste heat from a flue gas at a temperature of 127.degree.
C.
[0420] The specific work output and energy density of the CES
depend mainly on the efficiencies of the turbine .eta..sub.T and
the air liquefaction .eta..sub.A. The efficiency of the compressor
can also be important if the input air is compressed. The heat
exchangers play an important role in determining the overall
efficiency of the cycle. A higher temperature of waste heat and a
higher temperature of environment give a higher efficiency.
[0421] The CES system has a number of critical inventive steps,
including the recycling of waste cold as well as waste heat. These
specifically improve the overall work cycle against previous
systems designed using cryogenic liquid as the working fluid.
[0422] The CES system has the potential to achieve a better
performance over the existing energy storage systems in terms of
energy density, lifetime, capital cost and environmental impact and
is a competitive technology with respect to the output power and
duration, and the energy efficiency.
[0423] The CES system has the potential to harness low grade heat
and no obvious barriers to engineering. The system can be built
using existing technologies for the liquefaction plant, turbine,
heat exchanges, compressors, pumps, etc.
[0424] The majority of work in a CES is achieved by harnessing
energy attributed to the temperature difference between the cryogen
(.about.77K) and the ambient (.about.300K), whereas a standard
geothermal or waste heat energy system can only harness
temperatures above ambient (.about.300K).
Sample Models of CPS Engines
[0425] Five marine engine models have been prepared using the CPS.
These models are then compared against five known diesel engines.
The details of the five known industrial diesel engines are shown
in table 4.
[0426] CAT-3516 is a 78.1 litre 60.degree. V-type 16-cylinder
diesel engine. This engine is designed for medium transportation
boats with medium speeds. CAT-3126 is a 7.2 litre turbocharged
aftercooled in-line 6-cylinder engine adapted for small yachts. The
ST3 engine is an air cooled diesel engine form Lister Petter
company designed for narrow boats. The Cummins 6-cylinder T/C
diesel engine is used by a Thames river liner suitable for public
transport applications and pleasure cruising.
TABLE-US-00004 TABLE 4 CAT-3516 CAT-3126 ST3 Cummins (Caterpillar
(Caterpillar Ford air cooled 6-cylinder Marine Power Marine Power
Porbeangle (Lister T/C Co Ltd) Co Ltd) 6-cylinder Petter)
Riverliner Total Power 2525 bkW 261 bkW 77.6 bkW 25 bkW 522 kW
Speed 1800 rpm 2800 rpm Working 24 days 24 hrs 10 hrs 42 hrs 24
days Time Heat 0 kW 0 kW 0 kW 0 kW 0 kW Refrigeration 0 kW 0 kW 0
kW 0 kW 0 kW Air 0 kW 0 kW 0 kW 0 kW 0 kW Condition Work Output
2525 kW .261 kW 77.6 kW 25 kW 522 kW Fuel 617 litre/hr 68 litre/hr
21 litre/hr 6.9 litre/hr 128 litre/hr Consumption Fuel Tank 355.4
m.sup.3 1.5 m.sup.3 0.21 m.sup.3 0.29 m.sup.3 14 m.sup.3 Volume
Tank Side 7.1 m 1.2 m 0.6 m 0.7 m 2.4 m Length.sup.1 Boat Speed 8.5
m/s 14 m/s 6 m/s 3 m/s 6 m/s (~17 knots) (~28 knots) (~12 knots)
(~6 knots) (~12 knots) Cruising 17,626 km 1,210 km ~210 km ~483 km
~3,100 km Range .sup.1Assuming a cubic tank.
[0427] CPS Model 1 corresponds to the CAT-3516 and is suitable for
medium sized boats. As the CPS can provide a large quantity of
cold, Model 1 is particularly designed for transportation of
materials below sub-ambient conditions e.g. frozen meat and fish or
other products. Model 1 also makes use of the cooling air and heat
from the CPS for the occupants of the boat.
[0428] Models 2 to 4 correspond to the CAT-3126, the Ford Porbeagle
and the Lister Petter ST3 engine and are suitable for small yachts
or boats for which there is no need for large scale refrigeration,
or for cool air for air conditioning. The CPS system is used to
provide both propulsion and heat for use by the occupants of the
boat e.g. for heating.
[0429] Model 5 corresponds to the Cummins Riverliner. The CPS
system is used to provide propulsion, cooling air and heat for the
boat occupants and cold for freezing foods. Only a small part
(.about.10%) of the cold capacity of CPS is assumed for freezing
foods because the requirement for freezing food is much lower than
that of model 1 for transportation of materials under sub-ambient
conditions. However, a cruising range of only 60 miles (110 km) is
required as the Cummins Riverliner is designed to provide 12 return
journeys of 5 nautical miles per day.
[0430] The typical working conditions of all five models are
P.sub.2=200 bar, P.sub.1=8 bar, T.sub.0=300K,
.eta..sub.T.eta..sub.COM=.eta..sub.P=0.88, n=1.2, .gamma.=0.90, and
T.sub.df=5.0K. The overall performance of Models 1 to 5 under these
typical conditions is presented in table 5.
TABLE-US-00005 TABLE 5 Model M1 M2 M3 M4 M5 Total Power (kW) 2525
261 77.2 25 599.5 Working Time 24 days 24 hrs 10 hrs 42 hrs 5 hrs
Heat (kW) 169.6 22.0 6.5 2.1 45.2 Cold Refrigeration (kW)
962.0.sup.1 0 0 0 51.4.sup.2 Cold for Air 962.0.sup.1 0 0 0
256.8.sup.2 Condition (kW) Work output (kW) 1955.4 253.7 71.7 22.9
522 Energy Efficiency 59.4% 47.3% 47.3% 47.3% 52.8% Speed (m/s) 7.8
13.9 5.9 3.0 6.0 Cruising Range (km) 16180 1198 ~207 ~477 110 Fuel
Consumption 23264.4 3016.7 892.3 289.0 6210.5 (litre/hr) Efficiency
of 88.1% 99.0% 99.5% 98.2% 99.8% Dissipation Volume of Fuel
(m.sup.3) 13400.3 73.2 9.0 12.1 31.0 Length of Side (m) 23.7 4.2
2.1 2.3 3.1 Heat transfer 5514.6 734.5 217.3 70.4 1309.3
requirement (kW) Volume of heat 27.8 2.9 0.9 0.3 6.6 exchangers
(m.sup.3) Conservative Volume 111.1 11.6 3.4 1.1 26.4 (m.sup.3)
Length of Side (m) 4.8 2.3 1.5 1.0 3.0 .sup.1Assuming the
quantities of cold for refrigeration and air condition are the
same. .sup.2Assuming the quantities of cold for refrigeration and
1/5 of that for air condition.
[0431] For a given boat and a given power, the cruising speed
v.sub.k can be calculated by:
P o = .DELTA. 2 v k 3 3 C o ##EQU00116##
where P.sub.o, .DELTA., v.sub.k, C.sub.o are power (work) of the
engine, tonnage of the boat, cruising speed of the boat and a ship
geometry related coefficient, respectively. Assuming the Model 1
CPS powered boat has the same boat body, tonnage .DELTA., and
coefficient C.sub.o as the data used for the CAT-3516 engine, the
cruising speed, V.sub.k1, is calculated using:
v k 1 = v k _ 3516 W O 1 P o _ 3516 3 ##EQU00117##
where W.sub.O1 is the work output of model 1. The cruising range of
model 1 is therefore given by C.sub.r1=v.sub.k1O.sub.t1 where
O.sub.t1 is the maximum working time.
[0432] It can be seen that for the same total power, the work
output of model 1 CPS for propulsion is .about.22.6% lower than
that of CAT-3516, while the cruising speed and range decrease only
by .about.8%. Furthermore, model 1 CPS provides .about.169.6 kW
heat, 962.0 kW refrigeration cold and 962.0 kW cold for air
conditioning at the same time.
[0433] Similarly, the cruising speed and range for CPS models 2 to
4 powered boat can be obtained according to the data of CAT-3126.
The cruising speed and range for a CPS model 2 powered boat
are:
v k 2 = v k _ 3126 W O 2 P o _ 3126 3 ##EQU00118##
and C.sub.r2=V.sub.k2O.sub.t2. The cruising speed and range for a
CPS model 3 powered boat are:
v k 3 = v k _ 3126 W O 3 P o _ 3126 3 ##EQU00119##
and C.sub.r3=V.sub.k3P.sub.t3. The cruising speed and range for a
CPS model 4 powered boat are:
v k 4 = v k _ 3126 W O 4 P o _ 3126 3 ##EQU00120##
and C.sub.r4=V.sub.k4O.sub.t4. For the same total power, the work
outputs of models 2 to 4 for propulsion are .about.2.8% lower than
those of the corresponding diesel engine. However, models 2 to 4
can provide 22.0 kW, 6.5 kW and 2.1 kW heat at the same time,
respectively. It can be seen that the cruising speed and the range
of models 2 to 4 of the CPS are .about.99.0% of those of the
corresponding diesel engines.
[0434] Similarly, the cruising speed and range for a CPS model 5
powered boat can be obtained according to the data of the
Riverliner. The cruising speed and range for a CPS model 5 powered
boat are:
v k 5 = v k _ Riverliner W O 5 P o _ Riverline 3 ##EQU00121##
and C.sub.r5=V.sub.k5O.sub.t5. For the same work output for
propulsion, the CPS model 5 provides .about.45.2 kW heat, 256.8 kW
refrigeration cold and 51.4 kW cold for air conditioning although
the total power is 14.8% higher than that of the corresponding
diesel engine.
[0435] The flow rate of the fuel (liquid air) can be calculated
by:
F l = P o E D .rho. l ##EQU00122##
where F.sub.l, P.sub.o, E.sub.D, .rho..sub.l are flow rate of
liquid air, power of the engine, energy density of CPS and density
of liquid air, respectively. The volume of the fuel tank is
expressed as:
V l = F l O t E dis ##EQU00123##
where V.sub.l, O.sub.t, E.sub.dis are volume of liquid air,
operation time and efficiency of heat dissipation of the tank. If a
cubic tank is assumed, the length of each side, d, is
d = V l 3 . ##EQU00124##
[0436] The maximum heat transfer requirement has been analysed and
estimated above. For the CPS with a unit work output (1 kW), the
heat transfer requirement is: Q=2.184 kW. The size of the heat
transfers exchangers for a unit work output is: V=0.011 m.sup.3. If
a factor of safety is given as 4, the size of the heat transfer
exchangers for a unit work output would be 0.044 m.sup.3.
[0437] On the basis of above data, a conservative estimation of the
total volume of heat exchangers for models 1 to 5 CPS are listed in
table 5.
Comparison of CPS with Diesel Engines
[0438] Energy density and price: A comparison between tables 4 and
5 shows that the fuel consumptions of models 1 to 5 CPS are 37.70,
44.36, 42.08, 42.5 and 42.3 times those of the corresponding diesel
engines respectively. Therefore, the energy densities of models 1
to 5 are 1/37.70, 1/44.36, 1/42.08, 1/42.5 and 1/42.3 of those of
the corresponding diesel engines.
[0439] To compare the price of specific power of the eight engines,
the price of electricity is taken as Price_e=6 pence/kWh and that
of diesel as Price_d=90 pence/litre, the energy consumption to
produce 1 kg of liquid air is taken as 0.4 kWh (W. F. Castle.
2002). The price of the specific power (Price_p) of the four models
is calculated as: CAT-3516=22.0 p/kWh, M1=25.0 p/kWh, CAT-3126=23.4
p/kWh, M2=31.3 p/kWh, Ford Porbeagle=24.7 p/kWh, M3=31.3 p/kWh,
Lister Petter ST3=24.8 p/kWh, M4=31.3 p/kWh, Cummins=22.1 p/kWh,
and M5=28.8 p/kWh.
[0440] The price of specific power for the CPS models is comparable
to the corresponding diesel models. If model 1 CPS is used for
boats for transportation of frozen materials, the price of specific
power would be very competitive to the counterpart diesel
model.
[0441] If the energy consumption to produce 1 kg of liquid air is
taken as 0.3 kWh, then the price of specific power (Price_p) for
models 1 to 4 CPS become respectively 18.8, 23.4, 23.4, 23.4 and
21.6 p/kWh .
[0442] Energy Efficiency The comparison of well-to-wheel
efficiencies among the five models is shown in table 6. The CPS
data is based on 0.4 kWh to produce 1 kg liquid air. It can be seen
that the efficiency of model 1 CPS is similar to that of CAT-3516
and the efficiency of models 2 to 4 CPS using liquid air as fuel is
lower than that of the corresponding diesel engines. The medium
sized CPS boats have a higher efficiency when used for
transportation of frozen materials than the small yachts do because
the small yachts do not fully recover the cold. The efficiencies of
CPS models 1 to 4 shown in brackets is that if the consumption of
producing 1 kg liquid air is taken to be 0.3 kWh.
TABLE-US-00006 TABLE 6 Model CAT-3126 Ford Porbeagle Models CAT-
Model 1 Lister Petter 2 to 4 Cummins Model 3516 CPS ST3 CPS
Riverliner 5 CPS Fuel Diesel Liquid Diesel Liquid Diesel Liquid Air
Air Air Fuel 94% 51.6% 94% 51.6% 94% 51.6% production (68.8%)
(68.8%) (68.8%) efficiency Peak brake 38% 59.4% 38% 47.3% 38% 52.8%
engine efficiency or stack efficiency Part load 70% 70% 70% 70% 70%
70% efficiency factor Transmission 85% 80% 85% 90% 85% 80%
efficiency Weight factor .times. 100% 100% 100% 100% 100% 100% Idle
factor Total cycle 21% 18% 21% 15% 21% 16% efficiency (24%) (20%)
(21%)
[0443] Life time and capital cost: Since all major components of
the CPS are similar to the CES, the life time of a CPS system is
also estimated to be about 20 to 40 years. The life time of the
diesel engines is considered to be about 17 years. However, it is
believed that the life time of CPS is higher than that of diesel
engines because there is no combustion process at high temperatures
involved in CPS, and there is no strong friction between pistons
and cylinders.
[0444] It is believed that the CPS is competitive in terms of
capital cost because there is little special requirement in terms
of components. In addition, a refrigeration system is made obsolete
in the case of refrigeration transportation boats.
[0445] Influences of systems on the environment: Diesel engines
involve combustion of fossil fuels and hence lead to emission of
contaminates. CPS is a totally zero emission and environmentally
benign system. If liquid air is produced by renewable energy, the
CPS system would be a complete `Green` power system. Furthermore,
contaminates can be removed during liquefaction process, which
would help with mitigating the negative environmental issues
associated with burning of fossil fuels. Undesirable airborne
particulates can also be removed during production of liquid
air.
[0446] Accordingly, Cryogenic Propulsion System (CPS) using liquid
air can be used to provide combustion free and non-polluting
maritime transportation. CPS has a competitive performance against
the diesel engines in terms of energy price, energy efficiency,
life time and capital cost and impact on the environment. CPS can
have a higher efficiency if the cold energy is recovered for e.g.
on-boat refrigeration and air-conditioning.
[0447] It will of course be understood that the present invention
has been described by way of example, and that modifications of
detail can be made within the scope of the invention as defined by
the following claims.
* * * * *